Datasets:

pkuAI4M

/

AOPS

like 1

Follow

AI4M 16

[ "We set up a coordinate system, with the starting point of the car at the origin . At time $t$ , the car is at $\\left(\\frac 23t,0\\right)$ and the center of the storm is at $\\left(\\frac{t}{2}, 110 - \\frac{t}{2}\\right)$ . Using the distance formula,\n\\begin{eqnarray*} \\sqrt{\\left(\\frac{2}{3}t - \\frac 12t\\right)^2 + \\left(110-\\frac{t}{2}\\right)^2} &\\le& 51\\\\ \\frac{t^2}{36} + \\frac{t^2}{4} - 110t + 110^2 &\\le& 51^2\\\\ \\frac{5}{18}t^2 - 110t + 110^2 - 51^2 &\\le& 0\\\\ \\end{eqnarray*}\nNoting that $\\frac 12 \\left(t_1+t_2 \\right)$ is at the maximum point of the parabola, we can use $-\\frac{b}{2a} = \\frac{110}{2 \\cdot \\frac{5}{18}} = \\boxed{198}$", "First do the same process for assigning coordinates to the car. The car moves $\\frac{2}{3}$ miles per minute to the right, so the position starting from $(0,0)$ is $\\left(\\frac{2}{3}t, 0\\right)$\nTake the storm as circle. Given southeast movement, split the vector into component, getting position $\\left(\\frac{1}{2}t, 110 - \\frac{1}{2}t\\right)$ for the storm's center. This circle with radius 51 yields $\\left(x - \\frac{1}{2}t\\right)^2 + \\left(y -110 + \\frac{1}{2}t\\right)^2 = 51^2$\nNow substitute the car's coordinates into the circle's:\n$\\left(\\frac{2}{3}t - \\frac{1}{2}t\\right)^2 + \\left(-110 + \\frac{1}{2}t\\right)^2 = 51^2$\nSimplifying and then squaring:\n$\\left(\\frac{1}{6}t\\right)^2 + \\left(-110 + \\frac{1}{2}t\\right)^2 = 51^2$\n$\\frac{1}{36}t^2 + \\frac{1}{4}t^2 - 110t + 110^2$\nForming into a quadratic we get the following, then set equal to 0, since the first time the car hits the circumference of the storm is $t_{1}$ and the second is $t_{2}$\n$\\frac{5}{18}t^2 - 110t + 110^2 - 51^2 = 0$\nThe problem asks for sum of solutions divided by 2 so sum is equal to:\n$-\\frac{b}{a} = -\\frac{-110}{\\frac{5}{18}} = 110\\cdot{\\frac{18}{5}} = 396\\cdot{\\frac{1}{2}} = \\boxed{198}$", "We only need to know how the storm and car move relative to each other, so we can find this by subtracting the storm's movement vector from the car's. This gives the car's movement vector as $\\left(\\frac{1}{6}, \\frac{1}{2}\\right)$ . Labeling the car's starting position A, the storm center B, and the right triangle formed by AB with a right angle at B and the car's path, we get the following diagram, with AD as our desired length since D is the average of the points where the car enters and exits the storm.\n\n$AB = 110$ , so $CB = \\frac{110}{3}$ . The Pythagorean Theorem then gives $AC = \\frac{110\\sqrt{10}}{3}$ , and since $\\bigtriangleup ABC \\sim \\bigtriangleup ADB$ $AD = (AB)\\frac{AB}{AC} = 33\\sqrt{10}$ . The Pythagorean Theorem now gives the car's speed as $\\sqrt{\\frac{5}{18}}$ , and finally $\\frac{33\\sqrt{10}}{\\sqrt{\\frac{5}{18}}} = \\boxed{198}$" ]

[ "Since \\[\\mathrm{Distance}=\\mathrm{Speed}\\cdot\\mathrm{Time},\\] we seek the sum \\[5\\cdot1+12\\cdot1+19\\cdot1+26\\cdot1+\\cdots=5+12+19+26+\\cdots,\\] in which there are $30$ terms.\nThe last term is $5+7\\cdot(30-1)=208.$ Therefore, the requested sum is \\[5+12+19+26+\\cdots+208=\\frac{5+208}{2}\\cdot30=\\boxed{3195}.\\] ~MRENTHUSIASM", "The distance (in inches) traveled within each $1$ -second interval is: \\[5,5+1(7),5+2(7), \\dots , 5+29(7).\\] This is an arithmetic sequence so the total distance travelled, found by summing them up is: \\[\\text{number of terms} \\cdot \\text{average of terms} = \\text{number of terms} \\cdot \\dfrac{\\text{first term}+\\text{last term}}{2}.\\] Or, \\[30 \\cdot \\dfrac{5+5+29(7)}{2} = 15 \\cdot 213 = \\boxed{3195}.\\] ~BakedPotato66", "From the $30$ -term sum \\[5+12+19+26+\\cdots\\] in Solution 1, taking modulo $10$ gives \\[5+12+19+26+\\cdots \\equiv 3\\cdot(5+2+9+6+3+0+7+4+1+8) = 3\\cdot45\\equiv5 \\pmod{10}.\\] The only answer choices congruent to $5$ modulo $10$ are $\\textbf{(A)}$ and $\\textbf{(D)}.$ By a quick estimation, $\\textbf{(A)}$ is too small, leaving us with $\\boxed{3195}.$", "This problem can be solved by physics method. This method is perhaps the quickest too and shows the beauty of the problem. The average speed increases $7 \\ \\text{in/s}$ per second. So, the acceleration $a=7 \\ \\text{in/s\\textsuperscript{2}}.$ The average speed of the first second is $5 \\ \\text{in/s}.$ We can know the initial velocity $v_0=5-0.5\\cdot7=1.5.$ The displacement at $t=30$ is \\[s=\\frac{1}{2}at^2+v_0t=\\frac{1}{2}\\cdot7\\cdot30^2+1.5\\cdot30= \\boxed{3195}.\\] ~Bran_Qin" ]

[ "The base price of Michelle's cell phone plan is $20$ dollars. \nIf she sent $100$ text messages and it costs $5$ cents per text, then she must have spent $500$ cents for texting, or $5$ dollars. She talked for $30.5$ hours, but $30.5-30$ will give us the amount of time exceeded past 30 hours. $30.5-30=.5$ hours $=30$ minutes.\nSince the price for phone calls is $10$ cents per minute, the additional amount Michelle has to pay for phone calls is $30*10=300$ cents, or $3$ dollars.\nAdding $20+5+3$ dollars $\\boxed{28}$" ]

[ "The base price of Michelle's cell phone plan is $20$ dollars. \nIf she sent $100$ text messages and it costs $5$ cents per text, then she must have spent $500$ cents for texting, or $5$ dollars. She talked for $30.5$ hours, but $30.5-30$ will give us the amount of time exceeded past 30 hours. $30.5-30=.5$ hours $=30$ minutes.\nSince the price for phone calls is $10$ cents per minute, the additional amount Michelle has to pay for phone calls is $30*10=300$ cents, or $3$ dollars.\nAdding $20+5+3$ dollars $\\boxed{28}$" ]

[ "We can start at $200$ and work our way down to $1$ . We want to press the button that multiplies by $2$ the most, but since we are going down instead of up, we divide by $2$ instead. If we come across an odd number, then we will subtract that number by $1$ . Notice\nSince we've reached $1$ , it's clear that the answer should be $\\boxed{9}$ . Because we only subtracted $1$ when we had to, this is optimal. ~Roy2020" ]

[ "Iterating the condition $f(3x) = 3f(x)$ , we find that $f(x) = 3^kf\\left(\\frac{x}{3^k}\\right)$ for positive integers $k$ . We know the definition of $f(x)$ from $1 \\le x \\le 3$ , so we would like to express $f(2001) = 3^kf\\left(\\frac{2001}{3^k}\\right),\\ 1 \\le \\frac{2001}{3^k} \\le 3 \\Longrightarrow k = 6$ . Indeed,\n\\[f(2001) = 729\\left[1 - \\left| \\frac{2001}{729} - 2\\right|\\right] = 186.\\]\nWe now need the smallest $x$ such that $f(x) = 3^kf\\left(\\frac{x}{3^k}\\right) = 186$ . The range of $f(x),\\ 1 \\le x \\le 3$ , is $0 \\le f(x) \\le 1$ . So when $1 \\le \\frac{x}{3^k} \\le 3$ , we have $0 \\le f\\left(\\frac{x}{3^k}\\right) = \\frac{186}{3^k} \\le 1$ . Multiplying by $3^k$ $0 \\le 186 \\le 3^k$ , so the smallest value of $k$ is $k = 5$ . Then,\n\\[186 = {3^5}f\\left(\\frac{x}{3^5}\\right).\\]\nBecause we forced $1 \\le \\frac{x}{3^5} \\le 3$ , so\n\\[186 = {3^5}f\\left(\\frac{x}{3^5}\\right) = 243\\left[1 - \\left| \\frac{x}{243} - 2\\right|\\right] \\Longrightarrow x = \\pm 57 + 2 \\cdot 243.\\]\nWe want the smaller value of $x = \\boxed{429}$", "First, we start by graphing the function when $1\\leq{x}\\leq3$ , which consists of the lines $y=x-1$ and $y=3-x$ that intersect at $(2,1)$ . Similarly, using $f(3x)=3f(x)$ , we get a dilation of our initial figure by a factor of 3 for the next interval and so on.\nObserve that the intersection of two lines always has coordinates $(2y,y)$ where $y=3^a$ for some $a$ . First, we compute $f(2001)$ . The nearest intersection point is $(1458,729)$ when $a=7$ . Therefore, we can safely assume that $f(2001)$ is somewhere on the line with a slope of $-1$ that intersects at that nearest point. Using the fact that the slope of the line is $-1$ , we compute $f(2001)=729-543=186$ . However, we want the minimum value such that $f(x)=186$ and we see that there is another intersection point on the left which has a $y>186$ , namely $(486,243)$ . Therefore, we want the point that lies on the line with slope $1$ that intersects this point. Once again, since the slope of the line is $1$ , we get $x=486-57=\\boxed{429}$" ]

[ "Using the Pythagorean Theorem , the diagonal of the square $\\sqrt{(1.5)^2+(1.5)^2}=\\sqrt{4.5}>2$ . Because this is longer than $2$ ( length of the sides of two adjacent squares), the card can be placed like so, covering $12$ squares. $\\rightarrow \\boxed{12}$" ]

[ "Index the rows with $i = 1, 2, 3, ..., 13$ Index the columns with $j = 1, 2, 3, ..., 17$\nFor the first row number the cells $1, 2, 3, ..., 17$ For the second, $18, 19, ..., 34$ and so on\nSo the number in row = $i$ and column = $j$ is $f(i, j) = 17(i-1) + j = 17i + j - 17$\nSimilarly, numbering the same cells columnwise we\nfind the number in row = $i$ and column = $j$ is $g(i, j) = i + 13j - 13$\nSo we need to solve\n$f(i, j) = g(i, j)$\n$17i + j - 17 = i + 13j - 13$\n$16i = 4 + 12j$\n$4i = 1 + 3j$\n$i = (1 + 3j)/4$\nWe get $(i, j) = (1, 1), f(i, j) = g(i, j) = 1$\n$(i, j) = (4, 5), f(i, j) = g(i, j) = 56$\n$(i, j) = (7, 9), f(i, j) = g(i, j) = 111$\n$(i, j) = (10, 13), f(i, j) = g(i, j) = 166$\n$(i, j) = (13, 17), f(i, j) = g(i, j) = 221$\n$\\boxed{555}$ $555$" ]

[ "Corners have $5$ spots to go and there are $4$ corners, so $5 \\times 4=20$ .\nEdges have $3$ spots to go and there are $4$ sides so, $3 \\times 4=12$ . \nThat gives us $20+12=32$ spots to go into totally.\nSo $\\boxed{32}$ is the answer.\n~andliu766" ]

[ "Arranging eight cubes is the same as arranging the nine cubes first, and then removing the last cube. In other words, there is a one-to-one correspondence between every arrangement of nine cubes, and every actual valid arrangement. Thus, we initially get $9!$ . However, we have overcounted, because the red cubes can be permuted to have the same overall arrangement, and the same applies with the blue and green cubes. Thus, we have to divide by the $2!$ ways to arrange the red cubes, the $3!$ ways to arrange the blue cubes, and the $4!$ ways to arrange the green cubes. Thus we have $\\frac {9!} {2! \\cdot 3! \\cdot 4!} = \\boxed{1,260}$ different possible towers.", "We can divide the problem into three cases, each representing one cube to be excluded:\nCase 1 : The red cube is excluded. This gives us the problem of arranging one red cube, three blue cubes, and four green cubes. The number of possible arrangements is $\\frac{8!}{4!\\cdot3!}=280$ . Note that we do not need to multiply by the number of red cubes because there is no way to distinguish between the first red cube and the second.\nCase 2 : The blue cube is excluded. This gives us the problem of arranging two red cubes, two blue cubes, and four green cubes. The number of possible arrangements is $\\frac{8!}{2!\\cdot2!\\cdot4!}=420$\nCase 3 : The green cube is excluded. This gives us the problem of arranging two red cubes, three blue cubes, and three green cubes. The number of possible arrangements is $\\frac{8!}{2!\\cdot3!\\cdot3!}=560$\nAdding up the individual cases from above gives the answer as $280+420+560=\\boxed{1,260}$" ]

[ "The process of choosing a block can be represented by a generating function. Each choice we make can match the 'plastic medium red circle' in one of its qualities $(1)$ or differ from it in $k$ different ways $(kx)$ . Choosing the material is represented by the factor $(1+1x)$ , choosing the size by the factor $(1+2x)$ , etc: \\[(1+x)(1+2x)(1+3x)^2\\] Expanding out the first two factors and the square: \\[(1+3x+2x^2)(1+6x+9x^2)\\] By expanding further we can find the coefficient of $x^2$ , which represents the number of blocks differing from the original block in exactly two ways. We don't have to expand it completely, but choose the terms which will be multiplied together to result in a constant multiple of $x^2$ \\[1\\cdot9+3\\cdot6+2\\cdot1=\\boxed{29}\\]", "The amount of ways we can differ in the material, sizes, colors, and shapes category is 1, 2, 3, and 3 respectively (we take away one because we cannot choose the characteristic already chosen). We can choose to differ two of these characteristics, so our answer is $1\\cdot2 + 1\\cdot3 + 1\\cdot3 + 2\\cdot3 + 2\\cdot3 + 3\\cdot3 = \\boxed{29}$" ]

[ "$30$ days multiplied by $0.5$ gallons a day results in $15$ gallons of water loss.\nThe remaining water is $200-15=185=\\boxed{185}$" ]

[ "\nIn the triangle $ABC$ we have $AB = 1+4 = 5$ and $BC=4-1 = 3$ , thus by the Pythagorean theorem we have $AC=4$\nLet $r$ be the radius of the small circle, and let $s$ be the perpendicular distance from $S$ to $\\overline{AC}$ . Moreover, the small circle is tangent to both other circles, hence we have $SA=1+r$ and $SB=4+r$\nWe have $SA = \\sqrt{s^2 + (1-r)^2}$ and $SB=\\sqrt{(4-s)^2 + (4-r)^2}$ . Hence we get the following two equations:\n\\begin{align*} s^2 + (1-r)^2 & = (1+r)^2 \\\\ (4-s)^2 + (4-r)^2 & = (4+r)^2 \\end{align*}\nSimplifying both, we get\n\\begin{align*} s^2 & = 4r \\\\ (4-s)^2 & = 16r \\end{align*}\nAs in our case both $r$ and $s$ are positive, we can divide the second one by the first one to get $\\left( \\frac{4-s}s \\right)^2 = 4$\nNow there are two possibilities: either $\\frac{4-s}s=-2$ , or $\\frac{4-s}s=2$\nIn the first case clearly $s<0$ , which puts the center on the wrong side of $A$ , so this is not the correct case.\n(Note: This case corresponds to the other circle that is tangent to both given circles and the common tangent line. By coincidence, due to the $4:1$ ratio between radii of $A$ and $B$ , this circle turns out to have the same radius as circle $B$ , with center directly left of center $B$ , and tangent to $B$ directly above center $A$ .)\nThe second case solves to $s=\\frac 43$ . We then have $4r = s^2 = \\frac {16}9$ , hence $r = \\boxed{49}$", "The horizontal line is the equivalent of a circle of curvature $0$ , thus we can apply Descartes' Circle Formula\nThe four circles have curvatures $0, 1, \\frac 14$ , and $\\frac 1r$\nWe have $2\\left(0^2+1^2+\\frac {1}{4^2}+\\frac{1}{r^2}\\right)=\\left(0+1+\\frac 14+\\frac 1r\\right)^2$\nSimplifying, we get $\\frac{34}{16}+\\frac{2}{r^2}=\\frac{25}{16}+\\frac{5}{2r}+\\frac{1}{r^2}$\n\\[\\frac{1}{r^2}-\\frac{5}{2r}+\\frac{9}{16}=0\\] \\[\\frac{16}{r^2}-\\frac{40}{r}+9=0\\] \\[\\left(\\frac{4}{r}-9\\right)\\left(\\frac{4}{r}-1\\right)=0\\]\nObviously $r$ cannot equal $4$ , therefore $r = \\boxed{49}$" ]

[ "Let $O$ be the center of the circle, $\\overline{AB}$ be the chord, and $M$ be the midpoint of $\\overline{AB},$ as shown below. Note that $\\overline{OM}\\perp\\overline{AB}.$ Since $OM=AM=BM=5,$ we conclude that $\\triangle OMA$ and $\\triangle OMB$ are congruent isosceles right triangles. It follows that $r=5\\sqrt2,$ so the area of $\\odot O$ is $\\pi r^2=\\boxed{50}$" ]

[ "The equations of the two circles are $(x+10)^2+(y+4)^2=169$ and $(x-3)^2+(y-9)^2=65$ . Rearrange them to $(x+10)^2+(y+4)^2-169=0$ and $(x-3)^2+(y-9)^2-65=0$ , respectively. Their intersection points are where these two equations gain equality. The two points lie on the line with the equation $(x+10)^2+(y+4)^2-169=(x-3)^2+(y-9)^2-65$ . We can simplify this like the following. $(x+10)^2+(y+4)^2-169=(x-3)^2+(y-9)^2-65 \\rightarrow (x^2+20x+100)+(y^2+8y+16)-(x^2-6x+9)-(y^2-18y+81)=104 \\rightarrow 26x+26y+26=104 \\rightarrow 26x+26y=78 \\rightarrow x+y=3$ . Thus, $c = \\boxed{3}$", "Note the specificity of the radii, $13$ and $\\sqrt{65}$ , and that specificity is often deliberately added to simplify the solution to a problem.\nOne may recognize $13$ as the hypotenuse of the $\\text{5-12-13}$ right triangle and $\\sqrt{65}$ as the hypotenuse of the right triangle with legs $1$ and $8$ . We can suppose that the legs of these triangles connect the circles' centers to their intersection along the gridlines of the plane.\nIf we suspect that one of the intersections lies $12$ units to the right of and $5$ units above the center of the first circle, we find the point $(-10 + 12,-4 + 5) = (2,1)$ , which is in fact $1$ unit to the left of and $8$ units below the center of the second circle at $(3,9)$\nPlugging $(2,1)$ into $x + y$ gives us $c = 2 + 1 = \\boxed{3}$" ]

[ "Assuming we do not already know an inscribed angle is always half of its central angle, we will try a different approach. Since $O$ is the center, $OC$ and $OA$ are radii and they are congruent. Thus, $\\triangle COA$ is an isosceles triangle. Also, note that $\\angle COB$ and $\\angle COA$ are supplementary, then $\\angle COA = 180 - 50 = 130^{\\circ}$ . Since $\\triangle COA$ is isosceles, then $\\angle OCA \\cong \\angle OAC$ . They also sum to $50^{\\circ}$ , so each angle is $\\boxed{25}$" ]

[ "Note that $\\angle AOC = 180^\\circ - 50^\\circ = 130^\\circ$ . Because triangle $AOC$ is isosceles, $\\angle CAB = (180^\\circ - 130^\\circ)/2 = \\boxed{25}$" ]

[ "The probability that the chord doesn't intersect the triangle is $\\frac{11}{25}$ . The only way this can happen is if the two points are chosen on the same arc between two of the triangle vertices. The probability that a point is chosen on one of the arcs opposite one of the base angles is $\\frac{2x}{360}=\\frac{x}{180}$ (this comes from the Central Angle Theorem, which states that the central angle from two points on a circle is always twice the inscribed angle from those two points), and the probability that a point is chosen on the arc between the two base angles is $\\frac{180-2x}{180}$ . Therefore, we can write \\[2\\left(\\frac{x}{180}\\right)^2+\\left(\\frac{180-2x}{180}\\right)^2=\\frac{11}{25}\\] This simplifies to \\[x^2-120x+3024=0\\]\n(Note that the simplification is quite tedious)\nWhich factors as \\[(x-84)(x-36)=0\\] So $x=84, 36$ . The difference between these is $\\boxed{048}$", "Because we know that we have an isosceles triangle with angles of $x$ (and we know that x is an inscribed angle), that means that the arc that is intercepted by this angle is $2x$ . We form this same conclusion for the other angle $x$ , and $180-2x$ . Therefore we get $3$ arcs, namely, $2x$ $2x$ , and $360-4x$ . To have the chords intersect the triangle, we need the two points selected (to make a chord) to be on completely different arcs. An important idea to understand is that order matters in this case, so we have the equation $2$ $\\frac{2x}{360}$ $\\frac{2x}{360}$ $2$ $2$ $\\frac{2x}{360}$ $\\frac{360-4x}{360}$ $\\frac{14}{25}$ which using trivial algebra gives you $x^2-120x+3024$ and factoring gives you $(x-84)(x-36)$ and so your answer is $\\boxed{048}$ . \n~jske25", "After constructing the circumscribed circle, realize that the only time when the chord does not intersect the circle is when our $2$ points fall on only one arc formed by the sides of the triangle. Thus, lets call our isosceles triangle $ABC$ , where $AB=BC$ . Thus, the arcs formed by $BC$ and $AB$ can be called $a$ , and the arc formed by $AC$ is called $b$ . So, we can create the following system\n\\[2a+b=1\\] \\[2a^2+4ab=\\frac{14}{25}\\] \\[2a^2+b^2=\\frac{11}{25}\\]\nNotice we are denoting $a$ and $b$ as our probabilities, which we will be converting to degrees later. The 2 remaining systems can be calculated by using our rule about intersecting arcs and chords. So, after some hairy algebra we get:\n\\[a=\\frac{1}{5}\\] if \\[b=\\frac{3}{5}\\] \\[a=\\frac{7}{15}\\] if \\[b=\\frac{1}{15}\\]\nFrom here we find the absolute difference by doing $\\frac{7}{15}-\\frac{1}{5} = \\frac{4}{15}$ . Converting to degrees, since the angles of a triangle add up to $180^o$ , we find that $\\frac{4}{15} \\cdot 180 =\\boxed{048}$ , which is our answer." ]

[ "Let the side of the largest square be $x$ . It follows that the diameter of the inscribed circle is also $x$ . Therefore, the diagonal of the square inscribed inscribed in the circle is $x$ . The side length of the smaller square is $\\dfrac{x}{\\sqrt{2}}=\\dfrac{x\\sqrt{2}}{2}$ . Similarly, the diameter of the smaller inscribed circle is $\\dfrac{x\\sqrt{2}}{2}$ . Hence, its radius is $\\dfrac{x\\sqrt{2}}{4}$ . The area of this circle is $\\left(\\dfrac{x\\sqrt{2}}{4}\\right)^2\\pi=\\dfrac{2\\pi x^2}{16}=\\dfrac{x^2\\pi}{8}$ , and the area of the largest square is $x^2$ . The ratio of the areas is $\\dfrac{\\dfrac{x^2\\pi}{8}}{x^2}=\\frac{\\cancel{x^2}\\pi}{8}\\cdot\\frac{1}{\\cancel{x^2}}=\\boxed{8}$", "Let the radius of the smallest circle be $r$ . Then the side length of the smaller square is $2r$ . The radius of the larger circle is half the length of the diagonal of the smaller square, so it is $\\sqrt{2}r$ . Hence the largest square has sides of length $2\\sqrt{2}r$ . The ratio of the area of the smallest circle to the area of the largest square is therefore $\\frac{\\pi r^2}{\\left(2\\sqrt{2}r\\right)^2} =\\boxed{8}.$" ]

[ "We know that $A = sr$ , where $A$ is the triangle's area, $s$ its semiperimeter, and $r$ its inradius. Since this particular triangle is a right triangle (which we can verify by the Pythagorean theorem), the area is half of $8*15 = 120$ , and the semiperimeter is half of $8 + 15 + 17 = 40$ . Therefore, the inradius is $\\frac{120}{40} = 3$ , so our answer is $\\boxed{3}$ and we are done." ]

[ "The area of the shaded region is the area of the circle subtracted from the area of the rectangle.\nThe diameter of the circle is $1$ , so the radius is $1/2$ and the area is \\[(1/2)^2\\pi = \\pi /4.\\]\nThe rectangle obviously has area $2\\times 3= 6$ , so the area of the shaded region is $6-\\pi / 4$ .\nThis is closest to $5\\rightarrow \\boxed{5}$" ]

[ "\nDraw a square around the star figure. The side length of this square is $4$ , because the side length is the diameter of the circle. The square forms $4$ -quarter circles around the star figure. This is the equivalent of one large circle with radius $2$ , meaning that the total area of the quarter circles is $4\\pi$ . The area of the square is $16$ . Thus, the area of the star figure is $16 - 4\\pi$ . The area of the circle is $4\\pi$ . Taking the ratio of the two areas, we find the answer is $\\boxed{4}$" ]

[ "Note that the diameter of the circle is equal to the shorter side of the rectangle. Since the radius is $5$ , the diameter is $2\\cdot 5 = 10$ .\nSince the sides of the rectangle are in a $2:1$ ratio, the longer side has length $2\\cdot 10 = 20$ .\nTherefore the area is $20\\cdot 10 = 200$ or $\\boxed{200}$" ]

[ "If the radius is $5$ , then the width is $10$ , hence the length is $20$ $10\\times20= \\boxed{200}.$" ]

[ "Solution by e_power_pi_times_i\nThe area of the larger circle is $A_1 + A_2 = 9\\pi$ . Then $A_1 , 9\\pi-A_1 , 9\\pi$ are in an arithmetic progression. Thus $9\\pi-(9\\pi-A_1) = 9\\pi-A_1-A_1$ . This simplifies to $3A_1 = 9\\pi$ , or $A_1 = 3\\pi$ . The radius of the smaller circle is $\\boxed{3}$" ]

[ "Let $r$ be the radius of our circle. For it to be tangent to the positive $x$ and $y$ axes, we must have $C=(r,r)$ . For the circle to be externally tangent to the circle centered at $(3,0)$ with radius $1$ , the distance between $C$ and $(3,0)$ must be exactly $r+1$\nBy the Pythagorean theorem the distance between $(r,r)$ and $(3,0)$ is $\\sqrt{ (r-3)^2 + r^2 }$ , hence we get the equation $(r-3)^2 + r^2 = (r+1)^2$\nSimplifying, we obtain $r^2 - 8r + 8 = 0$ . By Vieta's formulas the sum of the two roots of this equation is $\\boxed{8}$" ]

[ "The formula for the area of a circle is $\\pi r^2$ so the radius of this circle is $\\sqrt{156}.$\nBecause $OA=4\\sqrt{3} < \\sqrt{156}, A$ must be in the interior of circle $O.$\nLet $s$ be the unknown value, the sidelength of the triangle, and let $X$ be the point on $BC$ where $OX \\perp BC.$ Since $\\triangle ABC$ is equilateral, $BX=\\frac{s}{2}$ and $AX=\\frac{s\\sqrt{3}}{2}.$ We are given $AO=4\\sqrt{3}.$ Use the Pythagorean Theorem and solve for $s.$\n\\begin{align*} (\\sqrt{156})^2 &= \\left(\\frac{s}{2}\\right)^2 + \\left( \\frac{s\\sqrt{3}}{2} + 4\\sqrt{3} \\right)^2\\\\ 156 &= \\frac14s^2 + \\frac34s^2 + 12s + 48\\\\ 0 &= s^2 + 12s - 108\\\\ 0 &= (s-6)(s+18)\\\\ s &= \\boxed{6}", "We can use the same diagram as Solution 1 and label the side length of $\\triangle ABC$ as $s$ . Using congruent triangles, namely the two triangles $\\triangle BOA$ and $\\triangle COA$ , we get that $\\angle BAO = \\angle CAO \\implies \\angle BAO = \\frac{360-60}{2} = 150$ . From this, we can use the Law of Cosines , to get \\[s^2 + (4 \\sqrt{3})^2 - 2 \\times s \\times 4 \\sqrt{3} \\times - \\frac{\\sqrt{3}}{2} = (2 \\sqrt{39})^2\\] Simplifying, we get \\[s^2 + 12s + 48 = 156 \\implies s^2 + 12s - 108 = 0\\] We can factor this to get \\[(x-6)(x+18)\\] Lengths must be non-negative, so the answer is $\\boxed{6}$ ~bryan gao" ]

[ "\nDraw radii $OE$ and $OD$ to the axes, and label the point of tangency to triangle $ABC$ point $F$ . Let the radius of the circle $O$ be $r$ . Square $OEAD$ has side length $r$\nBecause $BD$ and $BF$ are tangents from a common point $B$ $BD = BF$\n$AD = AB + BD$\n$r = 1 + BD$\n$r = 1 + BF$\nSimilarly, $CF = CE$ , and we can write:\n$AE = AC + CE$\n$r = \\sqrt{3} + CF$\nEquating the radii lengths, we have $1 + BF = \\sqrt{3} + CF$\nThis means $BF - CF = \\sqrt{3} - 1$\n$BF + CF = 2$ by the 30-60-90 triangle.\nTherefore, $2BF = 2 + \\sqrt{3} - 1$ , and we get $BF = \\frac{1}{2} + \\frac{\\sqrt{3}}{2}$\nThe radius of the circle is $AD$ , which is $BF + 1 = \\frac{3}{2} + \\frac{\\sqrt{3}}{2}$\nUsing decimal approximations, $r \\approx 1.5 + \\frac{1.73^+}{2} \\approx 2.37$ , and the answer is $\\boxed{2.37}$" ]

[ "1994 AIME Problem 2 - Solution.png\nCall the center of the larger circle $O$ . Extend the diameter $\\overline{PQ}$ to the other side of the square (at point $E$ ), and draw $\\overline{AO}$ . We now have a right triangle , with hypotenuse of length $20$ . Since $OQ = OP - PQ = 20 - 10 = 10$ , we know that $OE = AB - OQ = AB - 10$ . The other leg, $AE$ , is just $\\frac 12 AB$\nApply the Pythagorean Theorem\nThe quadratic formula shows that the answer is $\\frac{16 \\pm \\sqrt{16^2 + 4 \\cdot 240}}{2} = 8 \\pm \\sqrt{304}$ . Discard the negative root, so our answer is $8 + 304 = \\boxed{312}$" ]

[ "Case 1: The tangent and the origin are on the opposite sides of the circle.\nIn this case, $a, b > 2r$\nWe can easily prove that \\[a + b - 2 r = c . \\hspace{1cm} (1)\\]\nRecall that $c = \\sqrt{a^2 + b^2}$\nTaking square of (1) and reorganizing all terms, (1) is converted as \\[\\left( a - 2 r \\right) \\left( b - 2 r \\right) = 2 r^2 .\\]\nCase 2: The tangent and the origin are on the same sides of the circle.\nIn this case, $0 < a, b < r$\nWe can easily prove that \\[2 r - a - b = c . \\hspace{1cm} (2)\\]\nRecall that $c = \\sqrt{a^2 + b^2}$\nTaking square of (2) and reorganizing all terms, (2) is converted as \\[\\left( a - 2 r \\right) \\left( b - 2 r \\right) = 2 r^2 .\\]\nPutting both cases together, for given $r$ , we look for solutions of $a$ and $b$ satisfying \\[\\left( a - 2 r \\right) \\left( b - 2 r \\right) = 2 r^2 ,\\] with either $a, b > 2r$ or $0 < a, b < r$\nNow, we need to find the smallest $r$ , such that the number of feasible solutions of $(a, b)$ is at least 14.\nFor equation \\[uv = 2 r^2 ,\\] we observe that the R.H.S. is a not a perfect square. Thus, the number of positive $(u, v)$ is equal to the number of positive divisors of $2 r^2$\nSecond, for each feasible positive solution $(u, v)$ , its opposite $(-u, -v)$ is also a solution. However, $(u,v)$ corresponds to a feasible solution if $(a, b)$ with $a = u + 2r$ and $b = v + 2r$ , but $(-u, -v)$ may not lead to a feasible solution if $(a, b)$ with $a = 2 r - u$ and $b = 2 r - v$\nRecall that we are looking for $r$ that leads to at least 14 solutions.\nTherefore, the above observations imply that we must have $r$ , such that $2 r^2$ has least 7 positive divisors.\nFollowing this guidance, we find the smallest $r$ is 6. This leads to the following solutions:\n$\\left( a_1, b_1, c_1 \\right) = \\left( 3, 4, 5 \\right)$ $\\left( a_2, b_2, c_2 \\right) = \\left( 4, 3, 5 \\right)$\n$\\left( a_3, b_3, c_3 \\right) = \\left( 20, 21, 29 \\right)$ $\\left( a_4, b_4, c_4 \\right) = \\left( 21, 20, 29 \\right)$\n$\\left( a_5, b_5, c_5 \\right) = \\left( 18, 24, 30 \\right)$ $\\left( a_6, b_6, c_6 \\right) = \\left( 24, 18, 30 \\right)$\n$\\left( a_7, b_7, c_7 \\right) = \\left( 16, 30, 34 \\right)$ $\\left( a_8, b_8, c_8 \\right) = \\left( 30, 16, 34 \\right)$\n$\\left( a_9, b_9, c_9 \\right) = \\left( 15, 36, 39 \\right)$ $\\left( a_{10}, b_{10}, c_{10} \\right) = \\left( 36, 15, 39 \\right)$\n$\\left( a_{11}, b_{11}, c_{11} \\right) = \\left( 14, 48, 50 \\right)$ $\\left( a_{12}, b_{12}, c_{12} \\right) = \\left( 48, 14, 50 \\right)$\n$\\left( a_{13}, b_{13}, c_{13} \\right) = \\left( 13, 84, 85 \\right)$ $\\left( a_{14}, b_{14}, c_{14} \\right) = \\left( 84, 13, 85 \\right)$\nTherefore, $\\frac{c_{14}}{c_1} = \\boxed{17}$", "As $c_i$ is the length of the segment $(0,a_i)$ and $(b_i,0)$ $a_i^2+b_i^2=c_i^2$ . The equation for the line that passes through $(0,a_i)$ and $(b_i,0)$ is $a_ix+b_iy-a_ib_i=0$\nBy the point-line distance formula from point $(r,r)$ to line $a_ix+b_iy-a_ib_i=0$\n\\[r = \\frac{ |a_ir+b_ir-a_ib_i| }{ \\sqrt{a_i^2+b_i^2} }, \\quad c_i = \\frac{ |a_ir+b_ir-a_ib_i| }{r} = |a_i+b_i- \\frac{a_ib_i}{r}|\\]\n\\[\\because c_i \\quad \\text{is an integer}, \\quad \\therefore r|a_ib_i\\]\nTo find $c_1$ and $c_{14}$ for the smallest $r$ , we will list Pythagorean triples $(a_i,b_i,c_i)$ with relatively prime elements from the smallest. Notice that we only need to find $7$ triples with $a_i<b_i$ , the $7$ other triples will simply be $(b_i,a_i,c_i)$ $a_i$ will not equal $b_i$ because then $c_i = a_i \\cdot \\sqrt{2}$ , meaning that $a_i$ $b_i$ , and $c_i$ cannot all be integers.\nThe $7$ smallest triples are: $(3,4,5)$ $(5,12,13)$ $(7,24,25)$ $(8,15,17)$ $(11,60,61)$ $(12,35,37)$ $(13,84,85)$\nTherefore, $\\frac{c_{14}}{c_1} = \\frac{85}{5} = \\boxed{17}$" ]

[ "Suppose that with a pair $(a_i,b_i)$ the circle is an excircle. Then notice that the hypotenuse must be $(r-x)+(r-y)$ , so it must be the case that \\[a_i^2+b_i^2=(2r-a_i-b_i)^2.\\] Similarly, if with a pair $(a_i,b_i)$ the circle is an incircle, the hypotenuse must be $(x-r)+(y-r)$ , leading to the same equation.\nNotice that this equation can be simplified through SFFT to \\[(a_i-2r)(b_i-2r)=2r^2.\\] Thus, we want the smallest $r$ such that this equation has at least $14$ distinct pairs $(a_i,b_i)$ for which this holds. The obvious choice to check is $r=6$ . In this case, since $2r^2=2^3\\cdot 3^2$ has $12$ positive factors, we get $12$ pairs, and we get another two if the factors are $-8,-9$ or vice versa. One can check that for smaller values of $r$ , we do not even get close to $14$ possible pairs.\nWhen $r=6$ , the smallest possible $c$ -value is clearly when the factors are negative. When this occurs, $a_i=4, b_i=3$ (or vice versa), so the mimimal $c$ is $5$ . The largest possible $c$ -value occurs when the largest of $a_i$ and $b_i$ are maximized. This occurs when the factors are $72$ and $1$ , leading to $a_i=84, b_i=13$ (or vice-versa), leading to a maximal $c$ of $85$\nHence the answer is $\\frac{85}5=\\boxed{17}$" ]

[ "The area of the smaller square is one half of the product of its diagonals. Note that the distance from a corner of the smaller square to the center is equivalent to the circle's radius so the diagonal is equal to the diameter: $2 \\cdot 2 \\cdot \\frac{1}{2}=2.$\nThe circle's shaded area is the area of the smaller square subtracted from the area of the circle: $\\pi - 2.$\nIf you draw the diagonals of the smaller square, you will see that the larger square is split $4$ congruent half-shaded squares. The area between the squares is equal to the area of the smaller square: $2.$\nApproximating $\\pi$ to $3.14,$ the ratio of the circle's shaded area to the area between the two squares is about\n\\[\\frac{\\pi-2}{2} \\approx \\frac{3.14-2}{2} = \\frac{1.14}{2} \\approx \\boxed{12}\\]", "For the ratio of the circle's shaded area to the area between the squares to be $1,$ they would have to be approximately the same size. For any ratio larger than that, the circle's shaded area must be greater. However, we can clearly see that the circle's shaded area is part of the area between the squares, and is approximately $\\boxed{12}$", "Set the side length of the bigger square to be $8$ . \nThen the area of the big square is $8^2 =64$ and \nthe area of the small square $(4\\sqrt{2})^2 = 32$ . \nThe difference is $32$ . \nThe area of the circle is $4^2$ times $\\pi$ which is $16 \\pi$ or about $48$ . \nKnowing the area of the small square is $32$ $48-32$ is $16$ . \nThe area of the big square is $64$ . \nSo $32/64$ is $1/2$ , or $\\boxed{12}$" ]

[ "\n$r_1 = O_1A = 24$ $r_2 = O_2B = 6$ $AG = BO_2 = r_2 = 6$ $O_1G = r_1 - r_2 = 24 - 6 = 18$ $O_1O_2 = r_1 + r_2 = 30$\n$\\triangle O_2BD \\sim \\triangle O_1GO_2$ $\\frac{O_2D}{O_1O_2} = \\frac{BO_2}{GO_1}$ $\\frac{O_2D}{30} = \\frac{6}{18}$ $O_2D = 10$\n$CD = O_2D + r_2 = 10 + 6 = 16$\n$EF = 2EC = EA + EB = AB = GO_2 = \\sqrt{(O_1O_2)^2-(O_1G)^2} = \\sqrt{30^2-18^2} = 24$\n$DEF = \\frac12 \\cdot EF \\cdot CD = \\frac12 \\cdot 24 \\cdot 16 = \\boxed{192}$", "Let the center of the circle with radius $6$ be labeled $A$ and the center of the circle with radius $24$ be labeled $B$ . Drop perpendiculars on the same side of line $AB$ from $A$ and $B$ to each of the tangents at points $C$ and $D$ , respectively. Then, let line $AB$ intersect the two diagonal tangents at point $P$ . Since $\\triangle{APC} \\sim \\triangle{BPD}$ , we have \\[\\frac{AP}{AP+30}=\\frac14 \\implies AP=10.\\] Next, throw everything on a coordinate plane with $A=(0, 0)$ and $B = (30, 0)$ . Then, $P = (-10, 0)$ , and if $C = (x, y)$ , we have \\[(x+10)^2+y^2=64,\\] \\[x^2+y^2=36.\\] Combining these and solving, we get $(x, y)=\\left(-\\frac{18}5, \\frac{24}5\\right)$ . Notice now that $P$ $C$ , and the intersections of the lines $x=6$ (the vertical tangent) with the tangent containing these points are collinear, and thus every slope between a pair of points will have the same slope, which in this case is $\\frac{-\\frac{18}5+10}{\\frac{24}5}=\\frac34$ . Thus, the other two vertices of the desired triangle are $(6, 12)$ and $(6, -12)$ . By the Shoelace Formula, the area of a triangle with coordinates $(-10, 0)$ $(6, 12)$ , and $(6, -12)$ is \\[\\frac12|-120-0-72-72+0-120|=\\boxed{192}.\\]", "\nFirst, we want to find $O_2D$ . We know that $\\angle O_1AD = \\angle O_2BD = 90^{\\circ}$ , so by AA similarity, $\\triangle O_1AD \\sim \\triangle O_2BD$ . We want to find the length of $x$ , and using the similar triangles, we write an equation: $\\frac{30 + x}{4} = x$ . Solving, we get $x=10$ . Therefore, $CD = 10 + 6 = 16$ . Next, we find that using AA similarity, $\\triangle O_2BD \\sim \\triangle HO_2D \\sim \\triangle ECD$ and they are 3-4-5 triangles. We can quickly compute $EF = 2EC = 2 \\cdot \\left( \\frac{3}{4} \\cdot 16 \\right) = 2 \\cdot 12 = 24$ . Therefore, the area is $\\frac{1}{2} \\cdot 16 \\cdot 24 = \\boxed{192}$" ]

[ "Consider the upper right half of the grid, which consists of a $4\\times4$ section of the checkerboard and a quarter-circle of radius $4$ . We can draw this as a coordinate grid and shade in the complete squares. There are $8$ squares in the upper right corner, so there are $8 \\cdot 4 = \\boxed{32}$ whole squares in total." ]

[ "The class could send $25$ carnations and no roses, $22$ carnations and $2$ roses, $19$ carnations and $4$ roses, and so on, down to $1$ carnation and $16$ roses. There are 9 total possibilities (from 0 to 16 roses, incrementing by 2 at each step), $\\Rightarrow \\boxed{9}$" ]

[ "The cost of a rose is odd, hence we need an even number of roses. Let there be $2r$ roses for some $r\\geq 0$ . Then we have $50-3\\cdot 2r = 50-6r$ dollars left. We can always reach the sum exactly $50$ by buying $(50-6r)/2 = 25-3r$ carnations. Of course, the number of roses must be such that the number of carnations is non-negative. We get the inequality $25-3r \\geq 0$ which solves to $r\\leq 8 \\frac13$ $r$ must be an integer, so there are $\\boxed{9}$ possible values of $r$ , and each gives us one solution.", "Let $x$ and $y$ be the number of roses and carnations bought. The equation should be $3x+2y = 50$ . Since $50$ is an even number, the product of $3x$ must be even and smaller than $50$ . You can try nonnegative even integers for $x$ and you will end up with the numbers $0$ $2$ $4$ $6$ $8$ $10$ $12$ $14$ , and $16$ . There are $9$ numbers in total, so the answer is $\\boxed{9}$", "Let $r$ represent the number of roses, and let $c$ represent the number of carnations. Then, we get the linear Diophantine equation, $3r+2c=50$ .\nUsing the Euclidean algorithm, we get the initial solutions to be $r_0=50$ and $c_0=-50$ , meaning the complete solution will be, $r=50+\\frac{2}{\\gcd(2,3)}$ $k=50+2k$ $c=-50-\\frac{3}{\\gcd(2,3)}k=-50-3k$\nThe solution range for which both $r$ and $c$ are positive is $17$ $\\leq k$ $\\leq$ $25$ . There are $\\boxed{9}$ possible values for $k$" ]

[ "First, find how many chimes will have already happened before midnight (the beginning of the day) of $\\text{February 27, 2003}.$ $13$ half-hours have passed, and the number of chimes according to the hour is $1+2+3+\\cdots+12.$ The total number of chimes is $13+78=91.$\nEvery day, there will be $24$ half-hours and $2(1+2+3+\\cdots+12)$ chimes according to the arrow, resulting in $24+156=180$ total chimes.\nOn $\\text{February 26},$ the number of chimes that still need to occur is $2003-91=1912.$ $1912 \\div 180=10 \\text{R}112.$ Rounding up, it is $11$ days past $\\text{February 26},$ which is $\\boxed{9}$" ]

[ "Let $k$ be the number of women selected. Then, the number of men not selected is $11-(k-1)=12-k$ .\nNote that the sum of the number of women selected and the number of men not selected is constant at $12$ . Each combination of women selected and men not selected corresponds to a committee selection. Since choosing 12 individuals from the total of 23 would give $k$ women and $12-k$ men, the number of committee selections is $\\binom{23}{12}$ . \nThe answer is $\\boxed{081}$ .\n~awang11's sol", "We casework on the amount of men on the committee.\nIf there are no men in the committee, there are $\\dbinom{12}{1}$ ways to pick the women on the committee, for a total of $\\dbinom{11}{0} \\cdot \\dbinom{12}{1}$ . Notice that $\\dbinom{11}{0}$ is equal to $\\dbinom{11}{11}$ , so the case where no men are picked can be grouped with the case where all men are picked. When all men are picked, all women must also be picked, for a total of $\\dbinom{12}{12}$ . Therefore, these cases can be combined to \\[\\dbinom{11}{0} \\cdot \\left(\\dbinom{12}{1} + \\dbinom{12}{12}\\right)\\] Since $\\dbinom{12}{12} = \\dbinom{12}{0}$ , and $\\dbinom{12}{0} + \\dbinom{12}{1} = \\dbinom{13}{1}$ , we can further simplify this to \\[\\dbinom{11}{0} \\cdot \\dbinom{13}{1}\\]\nAll other cases proceed similarly. For example, the case with one men or ten men is equal to $\\dbinom{11}{1} \\cdot \\dbinom{13}{2}$ . Now, if we factor out a $13$ , then all cases except the first two have a factor of $121$ , so we can factor this out too to make our computation slightly easier. The first two cases (with $13$ factored out) give $1+66=67$ , and the rest gives $121(10+75+270+504) = 103,939$ . Adding the $67$ gives $104,006$ . Now, we can test for prime factors. We know there is a factor of $2$ , and the rest is $52,003$ . We can also factor out a $7$ , for $7,429$ , and the rest is $17 \\cdot 19 \\cdot 23$ . Adding up all the prime factors gives $2+7+13+17+19+23 = \\boxed{081}$" ]

[ "The probability of three heads and five tails is $\\binom {8}{3}p^3(1-p)^5$ and the probability of five heads and three tails is $\\binom {8}{3}p^5(1-p)^3$\n\\begin{align*} 25\\binom {8}{3}p^3(1-p)^5&=\\binom {8}{3}p^5(1-p)^3 \\\\ 25(1-p)^2&=p^2 \\\\ 25p^2-50p+25&=p^2 \\\\ 24p^2-50p+25&=0 \\\\ p&=\\frac {5}{6}\\end{align*}\nTherefore, the answer is $5+6=\\boxed{011}$", "We start as shown above. However, when we get to $25(1-p)^2=p^2$ , we square root both sides to get $5(1-p)=p$ . We can do this because we know that both $p$ and $1-p$ are between $0$ and $1$ , so they are both positive. Now, we have:\n\\begin{align*} 5(1-p)&=p \\\\ 5-5p&=p \\\\ 5&=6p \\\\ p&=\\frac {5}{6}\\end{align*}\nNow, we get $5+6=\\boxed{011}$", "Rewrite it as : $(P)^3$ $(1-P)^5=\\frac {1}{25}$ $(P)^5$ $(1-P)^3$\nThis can be simplified as $24P^2 -50P + 25 = 0$\nThis can be factored into $(4P-5)(6P-5)$\nThis yields two solutions: $\\frac54$ (ignored because it would result in $1-p<0$ ) or $\\frac56$\nTherefore, the answer is $5+6$ $\\boxed{011}$" ]

[ "We proceed recursively . Suppose we can build $T_m$ towers using blocks of size $1, 2, \\ldots, m$ . How many towers can we build using blocks of size $1, 2, \\ldots, m, m + 1$ ? If we remove the block of size $m + 1$ from such a tower (keeping all other blocks in order), we get a valid tower using blocks $1, 2, \\ldots, m$ . Given a tower using blocks $1, 2, \\ldots, m$ (with $m \\geq 2$ ), we can insert the block of size $m + 1$ in exactly 3 places: at the beginning, immediately following the block of size $m - 1$ or immediately following the block of size $m$ . Thus, there are 3 times as many towers using blocks of size $1, 2, \\ldots, m, m + 1$ as there are towers using only $1, 2, \\ldots, m$ . There are 2 towers which use blocks $1, 2$ , so there are $2\\cdot 3^6 = 1458$ towers using blocks $1, 2, \\ldots, 8$ , so the answer is $\\boxed{458}$" ]

[ "$2000\\%$ is equivalent to $20\\times100\\%$ . Therefore, $2000\\%$ of a number is the same as $20$ times that number. $4$ quarters is $1$ dollar, so Bryden will get $20\\times1={20}$ dollars, $\\boxed{20}$", "Since $2000\\%$ is just $\\frac{2000}{100}$ , we can multiply that by $100$ , because four quarters is a $100$ cents. After the multiplication, we get $2000$ . Since our answer is in cents right now, we need to convert it to dollars, which would be $\\boxed{20}$ dollars." ]

[ "The trip was $240$ miles long and took $\\dfrac{120}{30}+\\dfrac{120}{20}=4+6=10$ gallons. Therefore, the average mileage was $\\dfrac{240}{10}= \\boxed{24}$", "Alternatively, we can use the harmonic mean to get $\\frac{2}{\\frac{1}{20} + \\frac{1}{30}} = \\frac{2}{\\frac{1}{12}} = \\boxed{24}$" ]

[ "The trip was $240$ miles long and took $\\dfrac{120}{30}+\\dfrac{120}{20}=4+6=10$ gallons. Therefore, the average mileage was $\\dfrac{240}{10}= \\boxed{24}$", "Alternatively, we can use the harmonic mean to get $\\frac{2}{\\frac{1}{20} + \\frac{1}{30}} = \\frac{2}{\\frac{1}{12}} = \\boxed{24}$" ]

[ "When the diameter is increased by $25\\%$ , it is increased by $\\dfrac{5}{4}$ , so the area of the base is increased by $\\left(\\dfrac54\\right)^2=\\dfrac{25}{16}$\nTo keep the volume the same, the height must be $\\dfrac{1}{\\frac{25}{16}}=\\dfrac{16}{25}$ of the original height, which is a $36\\%$ reduction. $\\boxed{36}$" ]

[ "When the diameter is increased by $25\\%$ , it is increased by $\\dfrac{5}{4}$ , so the area of the base is increased by $\\left(\\dfrac54\\right)^2=\\dfrac{25}{16}$\nTo keep the volume the same, the height must be $\\dfrac{1}{\\frac{25}{16}}=\\dfrac{16}{25}$ of the original height, which is a $36\\%$ reduction. $\\boxed{36}$" ]

[ "There are $3600$ seconds per hour, so we have \\begin{align*} \\frac{3600\\text{ seconds}}{\\text{hour}}\\cdot \\frac{10,000\\text{ additions}}{\\text{second}} &= \\frac{36,000,000\\text{ additions}}{\\text{hour}} \\\\ &= 36\\text{ million additions per hour} \\end{align*}\n$\\boxed{36}$" ]

[ "The scale factor is uniform in all dimensions, so the volume of the liquid is $\\left(\\frac{3}{4}\\right)^{3}$ of the container. The remaining section of the volume is $\\frac{1-\\left(\\frac{3}{4}\\right)^{3}}{1}$ of the volume, and therefore $\\frac{\\left(1-\\left(\\frac{3}{4}\\right)^{3}\\right)^{1/3}}{1}$ of the height when the vertex is at the top.\nSo, the liquid occupies $\\frac{1-\\left(1-\\left(\\frac{3}{4}\\right)^{3}\\right)^{1/3}}{1}$ of the height, or $12-12\\left(1-\\left(\\frac{3}{4}\\right)^{3}\\right)^{1/3}=12-3\\left(37^{1/3}\\right)$ . Thus $m+n+p=\\boxed{052}$", "(Computational) The volume of a cone can be found by $V = \\frac{\\pi}{3}r^2h$ . In the second container, if we let $h',r'$ represent the height, radius (respectively) of the air (so $12 -h'$ is the height of the liquid), then the volume of the liquid can be found by $\\frac{\\pi}{3}r^2h - \\frac{\\pi}{3}(r')^2h'$\nBy similar triangles , we find that the dimensions of the liquid in the first cone to the entire cone is $\\frac{3}{4}$ , and that $r' = \\frac{rh'}{h}$ ; equating,\n\\begin{align*}\\frac{\\pi}{3}\\left(\\frac{3}{4}r\\right)^2 \\left(\\frac{3}{4}h\\right) &= \\frac{\\pi}{3}\\left(r^2h - \\left(\\frac{rh'}{h}\\right)^2h'\\right)\\\\ \\frac{37}{64}r^2h &= \\frac{r^2}{h^2}(h')^3 \\\\ h' &= \\sqrt[3]{\\frac{37}{64} \\cdot 12^3} = 3\\sqrt[3]{37}\\end{align*}\nThus the answer is $12 - h' = 12-3\\sqrt[3]{37}$ , and $m+n+p=\\boxed{052}$", "From the formula $V=\\frac{\\pi r^2h}{3}$ , we can find that the volume of the container is $100\\pi$ . The cone formed by the liquid is similar to the original, but scaled down by $\\frac{3}{4}$ in all directions, so its volume is $100\\pi*\\frac{27}{64}=\\frac{675\\pi}{16}$ . The volume of the air in the container is the volume of the container minus the volume of the liquid, which is $\\frac{925\\pi}{16}$ , which is $\\frac{37}{64}$ of the volume of the container. When the point faces upwards, the air forms a cone at the top of the container. This cone must have $\\sqrt[3]{\\frac{37}{64}}=\\frac{\\sqrt[3]{37}}{4}$ of the height of the container. This means that the height of the liquid is $12\\left(1-\\frac{\\sqrt[3]{37}}{4}\\right)=12-3\\sqrt[3]{37}$ inches, so our answer is $\\boxed{052}$ . Solution by Zeroman", "We find that the volume of the cone is $100\\pi$\nThe volume of the cone with height 9 is $\\frac{675}{16}\\pi$\nThe difference between the two volumes is $\\frac{925}{16}\\pi$ . Note that this is the volume of the cone essentially 'on top of' the frustum described in the problem when the liquid is held with the base horizontal.\nWe can express the volume as $x^2\\pi\\cdot\\frac{12}{5}x\\cdot\\frac{1}{3}$ , where x is the radius of the cone.\nSolving this equation, we get $x=\\frac{5\\sqrt[3]{37}}{4}$ . The height of this cone is $\\frac{12}{5}$ of the radius.\nThen we subtract the value from the height, 12, to get our answer: $12-3\\sqrt[3]{37}$\nTherefore, our answer is $12+3+37=\\boxed{52}$ . ~MC413551" ]

[ "Euler's Polyhedron Formula applied to $Q$ gives $n - 100 + F = 2$ , where F is the number of faces of $Q$ . Each edge of $Q$ is cut by two planes, so $R$ has $200$ vertices. Each cut by a plane $P_k$ creates an additional face on $R$ , so Euler's Polyhedron Formula applied to $R$ gives $200 - E + (F+n) = 2$ , where $E$ is the number of edges of $R$ . Subtracting the first equation from the second gives $300 - E = 0$ , whence $E = \\boxed{300}$", "Each edge connects two points. The plane cuts that edge so it splits into $2$ at each end (like two legs) for a total of $4$ new edges.\n\nBut because each new edge is shared by an adjacent original edge cut similarly, the additional edges are overcounted $\\times 2$\n\nSince there are $100$ edges to start with, $400/2=200$ new edges result. So there are $100+200=\\boxed{300}$ edges in the figure.", "The question specifies the slices create as many pyramids as there are vertices, implying each vertex owns 4 edge ends. There are twice as many edge-ends as there are edges, and $2 \\cdot 100 = 200$\n$\\frac{200}{4} = 50$ , so there are $50$ vertices.\nThe base of a pyramid has 4 edges, so each sliced vertex would add four edges to $R$\n$100 + 4 \\cdot 50$ $\\boxed{300}$", "I doubt anyone has the mental capacity to imagine $Q$ in its full complexity in their mind. So, we begin by examining a simpler structure: a cube.\nLet $E_C$ denote the number of edges of a polyhedron $C$ . Let $V_C$ denote the number of vertices of $C$ . Let $B_C$ denote the number of edges that meet at each vertex.\nSuppose $C$ is a cube. Then\n$E_C = 12$ $V_C = 8$ , and $B_C = 3$\n(A cube has $12$ edges, $8$ vertices, and $3$ edges meet per vertex)\nNotice that\n$V_C*B_C = 2E_C$ $8*3 = 2*12$\nWhy? Well, imagine an ant walking along an edge of a cube. When it reaches the end of the edge, it will be standing on a vertex (by definition). But, what if it had walked the other way? Well, it would still be on the same edge, but it would reach the other endpoint of the edge, another vertex. Therefore, we can say that a polyhedron with $E$ edges has $2E$ total endpoints. In the case of $C$ , a cube, this means a cube has $2*12 = 24$ total endpoints. But a cube only has $8$ vertices, not $24$ ... This doesn't make any sense... Wait a minute, we're dealing with a polyhedron, which means multiple edges will be connected to the same vertex. In the case of a cube, $3$ edges meet at every vertex. So we have $24 \\div 3$ , or $8$ vertices. The math works out. Let's apply this to $Q$ , our mystery polyhedron.\nWe are told that $E_Q = 100$ $V_Q = n$ and $B_Q$ isn't given explicitly. So we'll let $B_Q = b$ . Now our equation reads\n$bn = 200$\nWhat does this mean, exactly? Well $Q$ has $100$ edges, so it must has $200$ endpoints (endpoints, not vertices). So if we can figure out either $n$ , the number of vertices, or $b$ , the number of edges per vertex, we can figure out the other variable. Hmmmm... The problem doesn't give us any clues about either. How frustrating. Let's leave this alone and consider the other information given to us.\nWe are also told that \"the polyhedron is cut by planes $P_1, P_2,...,P_n$ \" (at this point I imagine $Q$ is a giant block of cheese, potato, etc.), and also that \"plane $P_k$ cuts only those edges that meet at $V_k$ \". What does this mean?\nWell, let's go back to our cube. Notice that there are exactly as many planes as vertices. Think of the planes as knives (or knife strokes) cutting into a big cube of cheese. We're cutting the cheese $8$ times, since there are $8$ vertices. Also, whenever we cut into the cheese, we have to direct the cuts near a vertex, cutting into only the edges connected to that vertex. How do we do that? Well, since we can't cut into any other edges, we're gonna have to settle for cutting corners (see what I did there? ...Ok, I admit it was cheesy... see what I did there, too?😂). We have to shave off just the corners of the cheese block, because this way it won't interfere with any of the other edges.\nThe problem also tells us that none of the planes overlap. In our cheese analogy, this means that every time we cut, we have to make sure the cuts are always into new cheese, and not into previously cut areas (all of the cuts are distinct, and don't overlap). Sure enough, this is in agreement with the information that exactly $n$ pyramids are cut off. Let's go back to $Q$\n$Q$ has $n$ vertices, so we're gonna make $n$ cuts. The question we want to answer is: \"After all of the cuts are made, how many total edges will $R$ (the figure that remains), have?\". Well, we're gonna have to figure out either $n$ or $b$ ... or do we? We're told that $n$ pyramids were made, but it doesn't specify exactly how many sides the pyramid has. So... that implies that it doesn't matter, or else the problem wouldn't be solvable. Let's suppose $b = 4$ $4$ edges meet at every vertex). Then we would have $200 \\div 4 = 50$ vertices, and we would make $50$ cuts. But how many more edges does it yield? (We are told that the planes don't overlap, so the original edges are all present.) Well, if we cut off the corner of a cube, it will create $3$ more edges... which is coincident with the fact that $3$ edges meet at every vertex. So for every vertex, we gain exactly $b$ new edges. Rewriting the equation to find the number of vertices, we have\n$v = \\frac{200}{b}$\nBut we just said that for every vertex/cut, we gain $b$ more edges. So the total number of new edges is $v*b \\Rightarrow \\frac{200}{b}*b \\Rightarrow 200$ . It doesn't matter how many vertices $Q$ has, or how many edges meet at every vertex. Given the conditions in the problem, we will always gain $200$ more edges, regardless of the values of $n$ , or $b$ . We have $200$ new edges $+$ $100$ old edges to make $300$ edges $\\Rightarrow \\boxed{300}$" ]

[ "Each edge of $Q$ is cut by two planes, so $R$ has $200$ vertices. Three edges of $R$ meet at each vertex, so $R$ has $\\frac 12 \\cdot 3 \\cdot 200 = \\boxed{300}$ edges.", "At each vertex, as many new edges are created by this process as there are original edges meeting at that vertex. Thus the total number of new edges is the total number of endpoints of the original edges, which is $200$ . A middle portion of each original edge is also present in $R$ , so $R$ has $100 + 200 = \\boxed{300}$ edges." ]

[ "Since we are asked to find $\\tan \\theta$ , we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$ . We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$ $B$ $C$ , and $D$ . Let $AB = 5$ $BC = 6$ $CD = 9$ , and $DA = 7$ . Let $AX = a$ $BX = b$ $CX = c$ , and $DX = d$ . We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$ We are given that the area of quadrilateral $ABCD$ is $30$ . We can express this area using the areas of triangles $AXB$ $BXC$ $CXD$ , and $DXA$ . Since we want to find $\\sin \\theta$ and $\\cos \\theta$ , we can represent these areas using $\\sin \\theta$ as follows: \\begin{align*} 30 &=[ABCD] \\\\ &=[AXB] + [BXC] + [CXD] + [DXA] \\\\ &=\\frac{1}{2} ab \\sin (\\angle AXB) + \\frac{1}{2} bc \\sin (\\angle BXC) + \\frac{1}{2} cd \\sin (\\angle CXD) + \\frac{1}{2} da \\sin (\\angle AXD) \\\\ &=\\frac{1}{2} ab \\sin (180^\\circ - \\theta) + \\frac{1}{2} bc \\sin (\\theta) + \\frac{1}{2} cd \\sin (180^\\circ - \\theta) + \\frac{1}{2} da \\sin (\\theta). \\end{align*} We know that $\\sin (180^\\circ - \\theta) = \\sin \\theta$ . Therefore it follows that: \\begin{align*} 30 &=\\frac{1}{2} ab \\sin (180^\\circ - \\theta) + \\frac{1}{2} bc \\sin (\\theta) + \\frac{1}{2} cd \\sin (180^\\circ - \\theta) + \\frac{1}{2} da \\sin (\\theta) \\\\ &=\\frac{1}{2} ab \\sin (\\theta) + \\frac{1}{2} bc \\sin (\\theta) + \\frac{1}{2} cd \\sin (\\theta) + \\frac{1}{2} da \\sin (\\theta) \\\\ &=\\frac{1}{2}\\sin\\theta (ab + bc + cd + da). \\end{align*} From here we see that $\\sin \\theta = \\frac{60}{ab + bc + cd + da}$ . Now we need to find $\\cos \\theta$ . Using the Law of Cosines on each of the four smaller triangles, we get following equations: \\begin{align*} 5^2 &= a^2 + b^2 - 2ab\\cos(180^\\circ-\\theta), \\\\ 6^2 &= b^2 + c^2 - 2bc\\cos \\theta, \\\\ 9^2 &= c^2 + d^2 - 2cd\\cos(180^\\circ-\\theta), \\\\ 7^2 &= d^2 + a^2 - 2da\\cos \\theta. \\end{align*} We know that $\\cos (180^\\circ - \\theta) = -\\cos \\theta$ for all $\\theta$ . We can substitute this value into our equations to get: \\begin{align*} 5^2 &= a^2 + b^2 + 2ab\\cos \\theta, &&(1) \\\\ 6^2 &= b^2 + c^2 - 2bc\\cos \\theta, &&(2) \\\\ 9^2 &= c^2 + d^2 + 2cd\\cos \\theta, &&(3) \\\\ 7^2 &= d^2 + a^2 - 2da\\cos \\theta. &&(4) \\end{align*} If we subtract $(2)+(4)$ from $(1)+(3)$ , the squared terms cancel, leaving us with: \\begin{align*} 5^2 + 9^2 - 6^2 - 7^2 &= 2ab \\cos \\theta + 2bc \\cos \\theta + 2cd \\cos \\theta + 2da \\cos \\theta \\\\ 21 &= 2\\cos \\theta (ab + bc + cd + da). \\end{align*} From here we see that $\\cos \\theta = \\frac{21/2}{ab + bc + cd + da}$\nSince we have figured out $\\sin \\theta$ and $\\cos \\theta$ , we can calculate $\\tan \\theta$ \\[\\tan \\theta = \\frac{\\sin \\theta}{\\cos \\theta} = \\frac{\\frac{60}{ab + bc + cd + da}}{\\frac{21/2}{ab + bc + cd + da}} = \\frac{60}{21/2} = \\frac{120}{21} = \\frac{40}{7}.\\] Therefore our answer is $40 + 7 = \\boxed{047}$", "In convex quadrilateral $ABCD,$ let $AB=5,BC=6,CD=9,$ and $DA=7.$ Let $A'$ and $C'$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to $\\overline{BD}.$ We obtain the following diagram: Let $BC'=p,C'E=q,EA'=r,A'D=s,AA'=h_1,$ and $CC'=h_2.$ We apply the Pythagorean Theorem to right triangles $\\triangle ABA',\\triangle BCC',\\triangle CDC',$ and $\\triangle DAA',$ respectively: \\[\\begin{array}{ccccccccccccccccc} (p+q+r)^2&+&h_1^2&=&5^2, &&&&&&&&&&&&\\hspace{36mm}(1) \\\\ [1ex] p^2&+&h_2^2&=&6^2, &&&&&&&&&&&&\\hspace{36mm}(2) \\\\ [1ex] (q+r+s)^2&+&h_2^2&=&9^2, &&&&&&&&&&&&\\hspace{36mm}(3) \\\\ [1ex] s^2&+&h_1^2&=&7^2. &&&&&&&&&&&&\\hspace{36mm}(4) \\end{array}\\] Let the brackets denote areas. We get \\begin{align*} [ABD]+[CBD]&=[ABCD] \\\\ \\frac12(p+q+r+s)h_1+\\frac12(p+q+r+s)h_2&=30 \\\\ \\frac12(p+q+r+s)(h_1+h_2)&=30 \\\\ (p+q+r+s)(h_1+h_2)&=60. \\hspace{49.25mm}(5) \\end{align*} We subtract $(2)+(4)$ from $(1)+(3):$ \\begin{align*} (p+q+r)^2+(q+r+s)^2-p^2-s^2&=21 \\\\ \\left[(p+q+r)^2-s^2\\right]+\\left[(q+r+s)^2-p^2\\right]&=21 \\\\ (p+q+r+s)(p+q+r-s)+(p+q+r+s)(-p+q+r+s)&=21 \\\\ (p+q+r+s)(2q+2r)&=21 \\\\ 2(p+q+r+s)(q+r)&=21 \\\\ (p+q+r+s)(q+r)&=\\frac{21}{2}. \\hspace{9.5mm}(6) \\end{align*} From right triangles $\\triangle AEA'$ and $\\triangle CEC',$ we have $\\tan\\theta=\\frac{h_1}{r}=\\frac{h_2}{q}.$ It follows that \\begin{alignat*}{8} \\tan\\theta&=\\frac{h_1}{r}\\qquad&\\implies\\qquad h_1&=r\\tan\\theta, \\hspace{64mm}&(1\\star)\\\\ \\tan\\theta&=\\frac{h_2}{q}\\qquad&\\implies\\qquad h_2&=q\\tan\\theta. &(2\\star) \\end{alignat*} Finally, we divide $(5)$ by $(6):$ \\begin{align*} \\frac{h_1+h_2}{q+r}&=\\frac{40}{7} \\\\ \\frac{r\\tan\\theta+q\\tan\\theta}{q+r}&=\\frac{40}{7} \\hspace{15mm} &&\\text{by }(1\\star)\\text{ and }(2\\star)\\\\ \\frac{(r+q)\\tan\\theta}{q+r}&=\\frac{40}{7} \\\\ \\tan\\theta&=\\frac{40}{7}, \\end{align*} from which the answer is $40+7=\\boxed{047}.$", "By Bretschneider's Formula, \\[30=\\tfrac{1}{4}\\sqrt{4u^2v^2-(b^2+d^2-a^2-c^2)^2}=\\tfrac{1}{4}\\sqrt{4u^2v^2-441}.\\] Thus, $uv=3\\sqrt{1649}$ . Also, \\[[ABCD]=\\tfrac 12 \\cdot uv\\sin{\\theta};\\] solving for $\\sin{\\theta}$ yields $\\sin{\\theta}=\\tfrac{40}{\\sqrt{1649}}$ . Since $\\theta$ is acute, $\\cos{\\theta}$ is positive, from which $\\cos{\\theta}=\\tfrac{7}{\\sqrt{1649}}$ . Solving for $\\tan{\\theta}$ yields \\[\\tan{\\theta}=\\frac{\\sin{\\theta}}{\\cos{\\theta}}=\\frac{40}{7},\\] for a final answer of $\\boxed{047}$" ]

[ "The same pattern is repeated for every $6 \\times 6$ tile. Looking closer, there is also symmetry of the top $3 \\times 3$ square, so the fraction of the entire floor in dark tiles is the same as the fraction in the square. Counting the tiles, there are $4$ dark tiles, and $9$ total tiles, giving a fraction of $\\boxed{49}$" ]

[ "Since $VYX + 1 = VVW$ , i.e. adding $1$ causes the \"fives\" digit to change, we must have $X = 4$ and $W = 0$ . Now since $VYZ + 1 = VYX$ , we have $X = Z + 1 \\implies Z = 4 - 1 = 3$ . Finally, note that in $VYX + 1 = VVW$ , adding $1$ will cause the \"fives\" digit to change by $1$ if it changes at all, so $V = Y + 1$ , and thus since $1$ and $2$ are the only digits left (we already know which letters are assigned to $0$ $3$ , and $4$ ), we must have $V = 2$ and $Y = 1$ . Thus $XYZ = 413_{5} = 4 \\cdot 5^{2} + 1 \\cdot 5 + 3 = 100 + 5 + 3 = 108$ , which is answer $\\boxed{108}$" ]

[ "The original cube has $6$ faces, each with an area of $2\\cdot 2 = 4$ square units. Thus the original figure had a total surface area of $24$ square units.\nThe new figure has the original surface, with $6$ new faces that each have an area of $1$ square unit, for a total surface area of of $6$ additional square units added to it. But $1$ square unit of the top of the bigger cube, and $1$ square unit on the bottom of smaller cube, is not on the surface, and does not count towards the surface area.\nThe total surface area is therefore $24 + 6 - 1 - 1 = 28$ square units.\nThe percent increase in surface area is $\\frac{SA_{new} - SA_{old}}{SA_{old}}\\cdot 100\\% = \\frac{28-24}{24}\\cdot 100\\% \\approx 16.67\\%$ , giving the closest answer as $\\boxed{17}$" ]

[ "On each face, there are $2$ diagonals like $x$ . There are $6$ faces on a cube. Thus, there are $2\\times 6 = 12$ diagonals that are \"x-like\".\nEvery \"y-like\" diagonal must connect the bottom of the cube to the top of the cube. Thus, for each of the $4$ bottom vertices of the cube, there is a different \"y-like\" diagonal. So there are $4$ \"y-like\" diagonals.\nThis gives a total of $12 + 4 = 16$ diagonals on the cube, which is answer $\\boxed{16}$", "There are $8$ vertices on a cube(A, B, C, D, E, F, G, H). If you pick one vertice, it will have 7 line segments associated with it, so you will have $\\frac{8\\cdot7}{2} = 28$ segments within the cube. The division by $2$ is necessary because you counted both the segment from $A$ to $B$ and the segment from $B$ to $A$\nBut not all $28$ of these segments are diagonals. Some are edges. There are $4$ edges on the top, $4$ edges on the bottom, and $4$ edges that connect the top to the bottom. So there are $12$ edges total, meaning that there are $28 - 12 = 16$ segments that are not edges. All of these segments are diagonals, and thus the answer is $\\boxed{16}$", "Consider picking one point on the corner of the cube. That point has $3$ \"x-like\" diagonals , and $1$ \"y-like\" diagonal that ends on the opposite vertex. Thus, each vertex has $3 + 1 = 4$ diagonals associated with it. There are $8$ vertices on the cube, giving a total of $8 \\cdot 4 = 32$ diagonals.\nHowever, each diagonal was counted as both a \"starting point\" and an \"ending point\". So there are really $\\frac{32}{2} = 16$ diagonals, giving an answer of $\\boxed{16}$" ]

[ "This problem is about the relationships between the white unit cubes and the blue unit cubes, which can be solved by Graph Theory . We use a Planar Graph to represent the larger cube. Each vertex of the planar graph represents a unit cube. Each edge of the planar graph represents a shared face between $2$ neighboring unit cubes. Each face of the planar graph represents a face of the larger cube.\nNow the problem becomes a Graph Coloring problem of how many ways to assign $4$ vertices blue and $4$ vertices white with Topological Equivalence . For example, in Figure $(1)$ , as long as the $4$ blue vertices belong to the same planar graph face, the different planar graphs are considered to be topological equivalent by rotating the larger cube.\nTopology.jpg\nHere is how the $4$ blue unit cubes are arranged:\nIn Figure $(1)$ $4$ blue unit cubes are on the same layer (horizontal or vertical).\nIn Figure $(2)$ $4$ blue unit cubes are in $T$ shape.\nIn Figure $(3)$ and $(4)$ $4$ blue unit cubes are in $S$ shape.\nIn Figure $(5)$ $3$ blue unit cubes are in $L$ shape, and the other is isolated without a shared face.\nIn Figure $(6)$ $2$ pairs of neighboring blue unit cubes are isolated from each other without a shared face.\nIn Figure $(7)$ $4$ blue unit cubes are isolated from each other without a shared face.\nSo the answer is $\\boxed{7}$", "Let’s split the cube into two layers; a bottom and top. Note that there must be four of each color, so however many number of one color are in the bottom, there will be four minus that number of the color on the top. We do casework on the color distribution of the bottom layer.\nCase 1: 4, 0\nIn this case, there is only one possibility for the top layer - all of the other color - $\\binom{4}{4}$ . Therefore there is 1 construction from this case.\nCase 2: 3, 1\nIn this case, the top layer has four possibilities, because there are four different ways to arrange it so that it also has a 3, 1 color distribution - $\\binom{4}{3}$ . Therefore there are 4 constructions from this case.\nCase 3: 2, 2\nIn this case, the top layer has six possibilities of arrangement - $\\binom{4}{2}$ . However, having adjacent colors one way can be rotated to having adjacent colors any other way, so there is only one construction for the adjacent colors subcase and similarly, only one for the diagonal color subcase. Therefore the total number of constructions for this case is 2.\nThe total number of constructions for the cube is thus $1+4+2=7=\\boxed{7}$", "Divide the $2 \\times 2 \\times 2$ cube into two layers, say, front and back. Any possible construction can be rotated such that the front layer has the same or greater number of white cubes than blue cubes, so we only need to count the number of cases given that is true.\nTherefore, our answer is $6+1+0=\\boxed{7}$", "Since rotations of a single pattern are considered indistinguishable, we can assume that the forward upper right corner of the 2-by-2-by-2 cube is a blue cube (since we can always rotate the big cube to place a blue cube in that spot).\nOnce we've assigned this cube to be blue, we note that 3 1-by-1-by-1 cubes share a side with it, 3 1-by-1-by-1 cubes share a corner with it, and 1 1-by-1-by-1 cube does not touch the assigned cube at all, from the perspective of someone who can only see the cube's faces. We'll call the first 3 \"adjacent\", the second 3 \"cornering\", and the last one \"opposite.\"\nWe can use a little bit of intuition to confirm that due to the rotation condition, we should treat all adjacents as indistinguishable, all cornerings as indistinguishable, and of course the opposite one is unique from all the others. Thus, we can list out like so (keeping in mind that there are 3 adjacents, 3 cornerings, and 1 opposite, and that we're choosing the positions of the remaining 3 blue cubes):\nOAA, OAC, OCC, CCC, CAA, CCA, AAA.\nThis gives the answer to be $\\boxed{7}$" ]

[ "Burnside lemma is used to counting number of orbit where the element on the same orbit can be achieved by the defined operator, naming rotation, reflection and etc.\nThe fact for Burnside lemma are\n1. the sum of stablizer on the same orbit equals to the # of operators;\n2. the sum of stablizer can be counted as $fix(g)$\n3. the sum of the $fix(g)/|G|$ equals the # of orbit.\nLet's start with defining the operator for a cube,\n1. $\\textbf{e (identity)}$\nFor identity, there are $\\frac{8!}{4!4!} = 70$\n2. ${\\bf r^{1}, r^{2}, r^{3}}$ to be the rotation axis along three pair of opposite face,\neach contains $r^{i}_{90}, r^{i}_{180}, r^{i}_{270}$ where $i= 1, 2, 3$\n$fix(r^{i}_{90}) = fix(r^{i}_{270}) = 2\\cdot1 = 2$\n$fix(r^{i}_{180}) = \\frac{4!}{2!\\cdot2!} = 6$\ntherefore $fix(\\bf r^{i}) = 2+2+6 = 10$ , and $fix(\\bf r^{1})+fix(\\bf r^{2})+fix(\\bf r^{3}) = 30$\n3. ${\\bf r^{4}, r^{5}, r^{6}, r^{7}}$ to the rotation axis along four cube diagnals.\neach contains $r^{i}_{120}, r^{i}_{240}$ where $i= 4, 5, 6, 7$\n$fix(r^{i}_{120}) = fix(r^{i}_{240}) = 2\\cdot1\\cdot2\\cdot1 = 4$\ntherefore $fix(\\bf r^{i}) = 4+4 = 8$ , and $fix(\\bf r^{4})+fix(\\bf r^{5})+fix(\\bf r^{6})+fix(\\bf r^{7}) = 32$\n4. ${\\bf r^{8}, r^{9}, r^{10}, r^{11}, r^{12}, r^{13}}$ to be the rotation axis along 6 pairs of diagnally opposite sides\neach contains $r^{i}_{180}$ where $i= 8, 9, 10, 11, 12, 13$\n$fix(r^{i}_{180}) = \\frac{4!}{2!\\cdot2!} = 6$\ntherefore $fix(\\bf r^{8})+fix(\\bf r^{9})+fix(\\bf r^{10})+fix(\\bf r^{11})+fix(\\bf r^{12})+fix(\\bf r^{13}) = 36$\n5. The total number of operators are\n$|G| = 1 + 3\\cdot3 + 4\\cdot2 + 6\\cdot1 = 24$\nBased on 1, 2, 3, 4 the total number of stablizer is $70 + 30 + 32 + 36 = 168$\ntherefore the number of orbit $= \\frac{168}{G=24} = \\boxed{7}$" ]

[ "This problem is about the relationships between the white unit cubes and the blue unit cubes, which can be solved by Graph Theory . We use a Planar Graph to represent the larger cube. Each vertex of the planar graph represents a unit cube. Each edge of the planar graph represents a shared face between $2$ neighboring unit cubes. Each face of the planar graph represents a face of the larger cube.\nNow the problem becomes a Graph Coloring problem of how many ways to assign $4$ vertices blue and $4$ vertices white with Topological Equivalence . For example, in Figure $(1)$ , as long as the $4$ blue vertices belong to the same planar graph face, the different planar graphs are considered to be topological equivalent by rotating the larger cube.\nTopology.jpg\nHere is how the $4$ blue unit cubes are arranged:\nIn Figure $(1)$ $4$ blue unit cubes are on the same layer (horizontal or vertical).\nIn Figure $(2)$ $4$ blue unit cubes are in $T$ shape.\nIn Figure $(3)$ and $(4)$ $4$ blue unit cubes are in $S$ shape.\nIn Figure $(5)$ $3$ blue unit cubes are in $L$ shape, and the other is isolated without a shared face.\nIn Figure $(6)$ $2$ pairs of neighboring blue unit cubes are isolated from each other without a shared face.\nIn Figure $(7)$ $4$ blue unit cubes are isolated from each other without a shared face.\nSo the answer is $\\boxed{7}$", "Let’s split the cube into two layers; a bottom and top. Note that there must be four of each color, so however many number of one color are in the bottom, there will be four minus that number of the color on the top. We do casework on the color distribution of the bottom layer.\nCase 1: 4, 0\nIn this case, there is only one possibility for the top layer - all of the other color - $\\binom{4}{4}$ . Therefore there is 1 construction from this case.\nCase 2: 3, 1\nIn this case, the top layer has four possibilities, because there are four different ways to arrange it so that it also has a 3, 1 color distribution - $\\binom{4}{3}$ . Therefore there are 4 constructions from this case.\nCase 3: 2, 2\nIn this case, the top layer has six possibilities of arrangement - $\\binom{4}{2}$ . However, having adjacent colors one way can be rotated to having adjacent colors any other way, so there is only one construction for the adjacent colors subcase and similarly, only one for the diagonal color subcase. Therefore the total number of constructions for this case is 2.\nThe total number of constructions for the cube is thus $1+4+2=7=\\boxed{7}$", "Divide the $2 \\times 2 \\times 2$ cube into two layers, say, front and back. Any possible construction can be rotated such that the front layer has the same or greater number of white cubes than blue cubes, so we only need to count the number of cases given that is true.\nTherefore, our answer is $6+1+0=\\boxed{7}$", "Since rotations of a single pattern are considered indistinguishable, we can assume that the forward upper right corner of the 2-by-2-by-2 cube is a blue cube (since we can always rotate the big cube to place a blue cube in that spot).\nOnce we've assigned this cube to be blue, we note that 3 1-by-1-by-1 cubes share a side with it, 3 1-by-1-by-1 cubes share a corner with it, and 1 1-by-1-by-1 cube does not touch the assigned cube at all, from the perspective of someone who can only see the cube's faces. We'll call the first 3 \"adjacent\", the second 3 \"cornering\", and the last one \"opposite.\"\nWe can use a little bit of intuition to confirm that due to the rotation condition, we should treat all adjacents as indistinguishable, all cornerings as indistinguishable, and of course the opposite one is unique from all the others. Thus, we can list out like so (keeping in mind that there are 3 adjacents, 3 cornerings, and 1 opposite, and that we're choosing the positions of the remaining 3 blue cubes):\nOAA, OAC, OCC, CCC, CAA, CCA, AAA.\nThis gives the answer to be $\\boxed{7}$" ]

[ "Burnside lemma is used to counting number of orbit where the element on the same orbit can be achieved by the defined operator, naming rotation, reflection and etc.\nThe fact for Burnside lemma are\n1. the sum of stablizer on the same orbit equals to the # of operators;\n2. the sum of stablizer can be counted as $fix(g)$\n3. the sum of the $fix(g)/|G|$ equals the # of orbit.\nLet's start with defining the operator for a cube,\n1. $\\textbf{e (identity)}$\nFor identity, there are $\\frac{8!}{4!4!} = 70$\n2. ${\\bf r^{1}, r^{2}, r^{3}}$ to be the rotation axis along three pair of opposite face,\neach contains $r^{i}_{90}, r^{i}_{180}, r^{i}_{270}$ where $i= 1, 2, 3$\n$fix(r^{i}_{90}) = fix(r^{i}_{270}) = 2\\cdot1 = 2$\n$fix(r^{i}_{180}) = \\frac{4!}{2!\\cdot2!} = 6$\ntherefore $fix(\\bf r^{i}) = 2+2+6 = 10$ , and $fix(\\bf r^{1})+fix(\\bf r^{2})+fix(\\bf r^{3}) = 30$\n3. ${\\bf r^{4}, r^{5}, r^{6}, r^{7}}$ to the rotation axis along four cube diagnals.\neach contains $r^{i}_{120}, r^{i}_{240}$ where $i= 4, 5, 6, 7$\n$fix(r^{i}_{120}) = fix(r^{i}_{240}) = 2\\cdot1\\cdot2\\cdot1 = 4$\ntherefore $fix(\\bf r^{i}) = 4+4 = 8$ , and $fix(\\bf r^{4})+fix(\\bf r^{5})+fix(\\bf r^{6})+fix(\\bf r^{7}) = 32$\n4. ${\\bf r^{8}, r^{9}, r^{10}, r^{11}, r^{12}, r^{13}}$ to be the rotation axis along 6 pairs of diagnally opposite sides\neach contains $r^{i}_{180}$ where $i= 8, 9, 10, 11, 12, 13$\n$fix(r^{i}_{180}) = \\frac{4!}{2!\\cdot2!} = 6$\ntherefore $fix(\\bf r^{8})+fix(\\bf r^{9})+fix(\\bf r^{10})+fix(\\bf r^{11})+fix(\\bf r^{12})+fix(\\bf r^{13}) = 36$\n5. The total number of operators are\n$|G| = 1 + 3\\cdot3 + 4\\cdot2 + 6\\cdot1 = 24$\nBased on 1, 2, 3, 4 the total number of stablizer is $70 + 30 + 32 + 36 = 168$\ntherefore the number of orbit $= \\frac{168}{G=24} = \\boxed{7}$" ]

[ "The cut $x = y$ separates the cube into points with $x < y$ and points with $x > y$ , and analogous results apply for the other cuts. Thus, which piece a particular point is in depends only on the relative sizes of its coordinates $x$ $y$ , and $z$ - for example, all points with the ordering $x < y < z$ are in the same piece. Thus, as there are $3! = 6$ possible orderings, there are $6$ pieces, which is answer $\\boxed{6}$" ]

[ "If none of the cubes have edge length $2$ , then all of the cubes have edge length $1$ , meaning they all are the same size, a contradiction.\nIt is clearly impossible to split a cube of edge $3$ into two or more cubes of edge $2$ with extra unit cubes, so there is one $2\\times 2\\times 2$ cube and $3^3-2^3=19$ unit cubes.\nThe total number of cubes, $N$ , is $1+19=20\\rightarrow \\boxed{20}$" ]

[ "Set the cube at the origin with the three vertices along the axes and the plane equal to $ax+by+cz+d=0$ , where $a^2+b^2+c^2=1$ . The distance from a point $(X,Y,Z)$ to a plane with equation $Ax+By+Cz+D=0$ is \\[\\frac{AX+BY+CZ+D}{\\sqrt{A^2+B^2+C^2}},\\] so the (directed) distance from any point $(x,y,z)$ to the plane is $ax+by+cz+d$ . So, by looking at the three vertices, we have $10a+d=10, 10b+d=11, 10c+d=12$ , and by rearranging and summing, \\[(10-d)^2+(11-d)^2+(12-d)^2= 100\\cdot(a^2+b^2+c^2)=100.\\]\nSolving the equation is easier if we substitute $11-d=y$ , to get $3y^2+2=100$ , or $y=\\sqrt {98/3}$ . The distance from the origin to the plane is simply $d$ , which is equal to $11-\\sqrt{98/3} =(33-\\sqrt{294})/3$ , so $33+294+3=\\boxed{330}$", "Let the vertices with distance $10,11,12$ be $B,C,D$ , respectively. An equilateral triangle $\\triangle BCD$ is formed with side length $10\\sqrt{2}$ . We care only about the $z$ coordinate: $B=10,C=11,D=12$ . It is well known that the centroid of a triangle is the average of the coordinates of its three vertices, so $\\text{centroid}=(10+11+12)/3=11$ . Designate the midpoint of $BD$ as $M$ . Notice that median $CM$ is parallel to the plane because the $\\text{centroid}$ and vertex $C$ have the same $z$ coordinate, $11$ , and the median contains $C$ and the $\\text{centroid}$ . We seek the angle $\\theta$ of the line: $(1)$ through the centroid $(2)$ perpendicular to the plane formed by $\\triangle BCD$ $(3)$ with the plane under the cube. Since the median is parallel to the plane, this orthogonal line is also perpendicular $\\textit{in slope}$ to $BD$ . Since $BD$ makes a $2-14-10\\sqrt{2}$ right triangle, the orthogonal line makes the same right triangle rotated $90^\\circ$ . Therefore, $\\sin\\theta=\\frac{14}{10\\sqrt{2}}=\\frac{7\\sqrt{2}}{10}$\nIt is also known that the centroid of $\\triangle BCD$ is a third of the way between vertex $A$ and $H$ , the vertex farthest from the plane. Since $AH$ is a diagonal of the cube, $AH=10\\sqrt{3}$ . So the distance from the $\\text{centroid}$ to $A$ is $10/\\sqrt{3}$ . So, the $\\Delta z$ from $A$ to the centroid is $\\frac{10}{\\sqrt{3}}\\sin\\theta=\\frac{10}{\\sqrt{3}}\\left(\\frac{7\\sqrt{2}}{10}\\right)=\\frac{7\\sqrt{6}}{3}$\nThus the distance from $A$ to the plane is $11-\\frac{7\\sqrt{6}}{3}=\\frac{33-7\\sqrt{6}}{3}=\\frac{33-\\sqrt{294}}{3}$ , and $33+294+3=\\boxed{330}$" ]

[ "\nLet's first view the cube from a direction perpendicular to $ABDC$ , as illustrated above. Let $x$ be the cube's side length. Since $\\triangle CHA \\sim \\triangle AGB$ , we have \\[\\frac{CA}{CH} = \\frac{AB}{AG}.\\] We know $AB = x$ $AG = \\sqrt{x^2-2^2}$ $AC = \\sqrt{2}x$ $CH = 8$ . Plug them into the above equation, we get \\[\\frac{\\sqrt{2}x}{8} = \\frac{x}{\\sqrt{x^2-2^2}}.\\] Solving this we get the cube's side length $x = 6$ , and $AC = 6\\sqrt{2}.$\nLet $PQ$ be the water's surface, both $P$ and $Q$ are $7$ meters from $\\mathcal P$ . Notice that $C$ is $8$ meters from $\\mathcal P$ , this means \\[CP = \\frac{1}{8}CA = \\frac{3\\sqrt{2}}{4}.\\] Similarly, \\[DQ = \\frac{3}{8}CA = \\frac{9\\sqrt{2}}{4}.\\]\n\nNow, we realize that the 3D space inside the cube without water is a frustum, with $P$ on its smaller base and $Q$ on its larger base. To find its volume, all we need is to find the areas of both bases and the height, which is $x = 6$ . To find the smaller base, let's move our viewpoint onto the plane $ABDC$ and view the cube from a direction parallel to $ABDC$ , as shown above. The area of the smaller base is simply \\[S_1 = CP^2 = \\Bigl(\\frac{3\\sqrt{2}}{4}\\Bigr)^2 = \\frac{9}{8}.\\] Similarly, the area of the larger base is \\[S_2 = DQ^2 = \\Bigl(\\frac{9\\sqrt{2}}{4}\\Bigr)^2 = \\frac{81}{8}.\\]\nFinally, applying the formula for a frustum's volume,\n\\[V = \\frac{1}{3} \\cdot x \\cdot (S_1 + \\sqrt{S_1S_2} + S_2) = \\frac{1}{3} \\cdot 6 \\cdot \\Bigl(\\frac{9}{8} + \\sqrt{\\frac{9}{8}\\cdot\\frac{81}{8}} + \\frac{81}{8}\\Bigl) = \\frac{117}{4}.\\]\nThe water's volume is thus \\[6^3 - \\frac{117}{4} = \\frac{747}{4},\\] giving $\\boxed{751}$", "Denote $h(X)$ the distance from point $X$ to $\\mathcal{P}, h(A) = 0, h(B) = 2,$ $h(C) = 8, h(D) = 10, h(G) = h(I) = h(H) = 7, AB = a, AC = a \\sqrt{2}.$\nLet slope $AB$ to $\\mathcal{P}$ be $\\alpha.$ Notation is shown in the diagram. \\[\\tan \\alpha = \\frac {\\sin \\alpha}{\\cos \\alpha} = \\frac {h(B)}{AB}\\cdot \\frac {AC}{h(C)} = \\frac {\\sqrt{2}}{4} \\implies a = 6.\\] Let $S = GI \\cap CD \\implies h(S) = h(G) = 7.$ \\[h(C) – h(G) = 8 - 7 = 1, h(D)- h(I) = 10 - 7 = 3.\\] \\[h(E) = h(F) = \\frac {h(D) +h(B)}{2} = 6 \\implies\\] \\[\\frac {DI}{DE} = \\frac {h(D) - h(I)}{h(D)-h(E)} = \\frac {3}{4} \\implies DI = DH = \\frac {9}{2}.\\]\nSimilarly $CG = \\frac {3}{2} \\implies SD = 9.$\nLet the volume without water be $V,$ volume of the pyramid $SCGJ$ be $U.$\nIt is clear that $U + V = 27U = \\frac {SD}{6} \\cdot DI^2 = \\frac {243}{8} \\implies$ $V = \\frac {243 \\cdot 26}{8 \\cdot 27 } = \\frac {117}{4} = 6^3 - \\frac {747}{4}$ from which $\\boxed{751}.$", "We introduce a Cartesian coordinate system to the diagram.\nWe put the origin at $A$ . We let the $z$ -components of $B$ $C$ $D$ be positive.\nWe set the $x$ -axis in a direction such that $B$ is on the $x-O-z$ plane.\nThe coordinates of $A$ $B$ $C$ are $A = \\left( 0, 0, 0 \\right)$ $B = \\left( x_B, 0 , 2 \\right)$ $C = \\left( x_C, y_C, 8 \\right)$\nBecause $AB \\perp AC$ $\\overrightarrow{AB} \\cdot \\overrightarrow{AC} = 0$ .\nThus, \\[ x_B x_C + 16 = 0 . \\hspace{1cm} (1) \\]\nBecause $AC$ is a diagonal of a face, $AC^2 = 2 AB^2$ .\nThus, \\[ x_C^2 + y_C^2 + 8^2 = 2 \\left( x_B^2 + 2^2 \\right) . \\hspace{1cm} (2) \\]\nBecause plane $ABCD$ is perpendicular to plan $P$ $\\hat z \\cdot \\left( \\overrightarrow{AB} \\times \\overrightarrow{AC} \\right) = 0$ .\nThus, \\[ \\begin{vmatrix} 0 & 0 & 1 \\\\ x_B & 0 & 2 \\\\ x_C & y_C & 8 \\end{vmatrix} = 0 . \\hspace{1cm} (3) \\]\nJointly solving (1), (2), (3), we get one solution $x_B = 4 \\sqrt{2}$ $x_C = - 2 \\sqrt{2}$ $y_C = 0$ .\nThus, the side length of the cube is $|AB| = \\sqrt{x_B^2 + 2^2} = 6$\nDenote by $P$ and $Q$ two vertices such that $AP$ and $AQ$ are two edges, and satisfy the right-hand rule that $\\widehat{AB} \\times \\widehat{AP} = \\widehat{AQ}$ .\nNow, we compute the coordinates of $P$ and $Q$\nBecause $|AB| = 6$ , we have $\\overrightarrow{AP} \\times \\overrightarrow{AQ} = 6 \\overrightarrow{AB}$ $\\overrightarrow{AQ} \\times \\overrightarrow{AB} = 6 \\overrightarrow{AP}$ $\\overrightarrow{AB} \\times \\overrightarrow{AP} = 6 \\overrightarrow{AQ}$\nHence, \\begin{align*} \\begin{bmatrix} \\hat i & \\hat j & \\hat k \\\\ x_P & y_P & z_P \\\\ x_Q & y_Q & z_Q \\end{bmatrix} & = 6 \\left( 4 \\sqrt{2} \\hat i + 2 \\hat k \\right) , \\\\ \\begin{vmatrix} \\hat i & \\hat j & \\hat k \\\\ x_Q & y_Q & z_Q \\\\ 4 \\sqrt{2} & 0 & 2 \\end{vmatrix} & = 6 \\left( x_P \\hat i + y_P \\hat j + z_P \\hat k \\right) , \\\\ \\begin{vmatrix} \\hat i & \\hat j & \\hat k \\\\ 4 \\sqrt{2} & 0 & 2 \\\\ x_P & y_P & z_P \\end{vmatrix} & = 6 \\left( x_Q \\hat i + y_Q \\hat j + z_Q \\hat k \\right) . \\end{align*}\nBy solving these equations, we get $y_P^2 + y_Q^2 = 36 .$\nIn addition, we have $\\overrightarrow{AC} = \\overrightarrow{AP} + \\overrightarrow{AQ}$ .\nThus, $P = \\left( - \\sqrt{2} , 3 \\sqrt{2} , 4 \\right)$ $Q = \\left( - \\sqrt{2} , - 3 \\sqrt{2} , 4 \\right)$\nTherefore, the volume of the water is \\begin{align*} V = & 6^3 \\int_{u=0}^1 \\int_{v=0}^1 \\int_{w=0}^1 \\mathbf 1 \\left\\{ z_B u + z_P v + z_Q w \\leq 7 \\right\\} dw dv du \\\\ & = 6^3 \\int_{u=0}^1 \\int_{v=0}^1 \\int_{w=0}^1 \\mathbf 1 \\left\\{ 2 u + 4 v + 4 w \\leq 7 \\right\\} dw dv du \\\\ & = 6^3 - 6^3 \\int_{u=0}^1 \\int_{v=0}^1 \\int_{w=0}^1 \\mathbf 1 \\left\\{ 2 u + 4 v + 4 w > 7 \\right\\} dw dv du . \\end{align*}\nDefine $u' = 1 - u$ $v' = 1 - v$ $w' = 1 - w$ .\nThus, \\begin{align*} V & = 6^3 - 6^3 \\int_{u=0}^1 \\int_{v=0}^1 \\int_{w=0}^1 \\mathbf 1 \\left\\{ 2 u' + 4 v' + 4 w' < 3 \\right\\} dw dv du \\\\ & = 6^3 - 6^3 \\int_{u'=0}^1 \\left( \\int_{v'=0}^1 \\int_{w'=0}^1 \\mathbf 1 \\left\\{ v' + w' < \\frac{3}{4} - \\frac{u'}{2} \\right\\} dw' dv' \\right) du' \\\\ & = 6^3 - 6^3 \\int_{u'=0}^1 \\frac{1}{2} \\left( \\frac{3}{4} - \\frac{u'}{2} \\right)^2 du' . \\end{align*}\nDefine $u'' = \\frac{3}{4} - \\frac{u'}{2}$ .\nThus, \\begin{align*} V & = 6^3 - 6^3 \\int_{u'' = 1/4}^{3/4} \\left( u'' \\right)^2 du'' \\\\ & = 6^3 - 6^3 \\frac{1}{3} \\left( \\left(\\frac{3}{4}\\right)^3 - \\left(\\frac{1}{4}\\right)^3 \\right) \\\\ & = 216 - \\frac{117}{4} \\\\ & = \\frac{747}{4} . \\end{align*}\nTherefore, the answer is $747 + 4 = \\boxed{751}$" ]

[ "Initially, the difference between the water temperature and the room temperature is $212-68=144$ degrees Fahrenheit.\nAfter $5$ minutes, the difference between the temperatures is $144\\div2=72$ degrees Fahrenheit.\nAfter $10$ minutes, the difference between the temperatures is $72\\div2=36$ degrees Fahrenheit.\nAfter $15$ minutes, the difference between the temperatures is $36\\div2=18$ degrees Fahrenheit. At this point, the water temperature is $68+18=\\boxed{86}$ degrees Fahrenheit." ]

[ "Our aim is to find the volume of the part of the cube submerged in the cylinder. \nIn the problem, since three edges emanate from each vertex, the boundary of the cylinder touches the cube at three points. Because the space diagonal of the cube is vertical, by the symmetry of the cube, the three points form an equilateral triangle. Because the radius of the circle is $4$ , by the Law of Cosines, the side length s of the equilateral triangle is\n\\[s^2 = 2\\cdot(4^2) - 2\\cdot(4^2)\\cos(120^{\\circ}) = 3(4^2)\\]\nso $s = 4\\sqrt{3}$ .* Again by the symmetry of the cube, the volume we want to find is the volume of a tetrahedron with right angles on all faces at the submerged vertex, so since the lengths of the legs of the tetrahedron are $\\frac{4\\sqrt{3}}{\\sqrt{2}} = 2\\sqrt{6}$ (the three triangular faces touching the submerged vertex are all $45-45-90$ triangles) so\n\\[v = \\frac{1}{3}(2\\sqrt{6})\\left(\\frac{1}{2} \\cdot (2\\sqrt{6})^2\\right) = \\frac{1}{6} \\cdot 48\\sqrt{6} = 8\\sqrt{6}\\]\nso\n\\[v^2 = 64 \\cdot 6 = \\boxed{384}.\\]", "Visualizing the corner which is submerged in the cylinder, we can see that it's like slicing off a corner, where they slice an equal part off every edge, to make a tetrahedron where there are 3 right isosceles triangles and one equilateral triangle. With this out of the way, we can now just find the area of that equilateral triangle, using the fact that the circle of radius $4$ is the circumcircle of the equilateral triangle. Using equilateral triangle properties, you can find that the height of the triangle is $6$ , and the side length is $\\frac{6}{\\sqrt{3}}=2\\sqrt{3} \\cdot 2=4\\sqrt{3}$ . As the other faces are right isosceles triangles, they are $\\frac{4\\sqrt{3}}{\\sqrt{2}}=2\\sqrt{6}$ . Therefore the volume of this tetrahedron is\n\\[\\left(\\frac{2\\sqrt{6}}{2}\\right)^2=12 \\ \\cdot \\ (2\\sqrt{6})=24\\sqrt{6} \\implies \\frac{24\\sqrt{6}}{3}=8\\sqrt{6} \\implies (8\\sqrt{6})^2=\\boxed{384}\\]", "We can use the same method as in Solution 2 to find the side length of the equilateral triangle, which is $4\\sqrt3$ . From here, its area is \\[\\dfrac{\\bigl(4\\sqrt3\\bigr)^2\\sqrt3}4=12\\sqrt3.\\] The leg of the isosceles right triangle is $\\dfrac{4\\sqrt3}{\\sqrt2}=2\\sqrt6$ , and the horizontal distance from the vertex to the base of the tetrahedron is $4$ (the radius of the cylinder), so we can find the height, as shown in the diagram. The height from the tip to the base is $2\\sqrt2$ , so the volume is $\\dfrac{12\\sqrt3\\cdot2\\sqrt2}3=8\\sqrt6$ , and thus the answer is $\\boxed{384}$", "Since the diagonal is perpendicular to the base of the cylinder, all three edges and faces can be treated symmetrically.\nThe cross-section of the cube with the top face of the cylinder is an equilateral triangle inscribed in a circle with radius $4$ . This means the medians of the triangle have length $\\frac{3}{2} \\cdot 4=6$ , because the circumcenter is also the centroid, and the centroid divides the medians into lengths of ratio $2:1$ . Using $30-60-90$ triangles, the side length of the triangle is $4\\sqrt{3}$ , and its area is $\\frac{(4\\sqrt{3})^2\\sqrt{3}}{4}=12\\sqrt{3}$\nNext, consider the submerged triangular sections of the faces. Each is a $45-45-90$ triangle with leg length $x$ . The area of each is then $\\frac{x^2}{2}$ . By De Gua's Theorem on the submerged pyramid (which we can apply because it has a right-angled corner), $3\\left( \\frac{x^2}{2} \\right) ^2=(12\\sqrt{3})^2$ . Solving yields $x=2\\sqrt{6}$\nThe height of the pyramid is then $\\sqrt{(2\\sqrt{6})^2-4^2}=2\\sqrt{2}$ , by the Pythagorean Theorem (using the slant height and circumradius). The volume is then $v=\\frac{1}{3}\\cdot 12\\sqrt{3} \\cdot 2\\sqrt{2}=8\\sqrt{6}$ , and the requested answer is $v^2=(8\\sqrt{6})^2=\\boxed{384}$ . ~bad_at_mathcounts" ]

[ "The volume of the wedge is half the volume of a cylinder with height $12$ and radius $6$ . (Imagine taking another identical wedge and sticking it to the existing one). Thus, $V=\\dfrac{6^2\\cdot 12\\pi}{2}=216\\pi$ , so $n=\\boxed{216}$" ]

[ "First, note that $1+7+5+2+5=20$ . There are $3$ possible cases:\nCase 1: the mean is $5$\n$X = 5 \\cdot 6 - 20 = 10$\nCase 2: the mean is $7$\n$X = 7 \\cdot 6 - 20 = 22$\nCase 3: the mean is $X$\n$X= \\frac{20+X}{6} \\Rightarrow X=4$\nTherefore, the answer is $10+22+4=\\boxed{36}$" ]

[ "First, note that $1+7+5+2+5=20$ . There are $3$ possible cases:\nCase 1: the mean is $5$\n$X = 5 \\cdot 6 - 20 = 10$\nCase 2: the mean is $7$\n$X = 7 \\cdot 6 - 20 = 22$\nCase 3: the mean is $X$\n$X= \\frac{20+X}{6} \\Rightarrow X=4$\nTherefore, the answer is $10+22+4=\\boxed{36}$" ]

[ "If the probability of choosing a red card is $\\frac{1}{3}$ , the red and black cards are in ratio $1:2$ . This means at the beginning there are $x$ red cards and $2x$ black cards.\nAfter $4$ black cards are added, there are $2x+4$ black cards. This time, the probability of choosing a red card is $\\frac{1}{4}$ so the ratio of red to black cards is $1:3$ . This means in the new deck the number of black cards is also $3x$ for the same $x$ red cards.\nSo, $3x = 2x + 4$ and $x=4$ meaning there are $4$ red cards in the deck at the start and $2(4) = 8$ black cards.\nSo, the answer is $8+4 = 12 = \\boxed{12}$", "In terms of the number of cards, the original deck is $3$ times the red cards, and the final deck is $4$ times the red cards. So, the final deck is $\\frac43$ times the original deck. We are given that adding $4$ cards to the original deck is the same as increasing the original deck by $\\frac13$ of itself. Since $4$ cards are equal to $\\frac13$ of the original deck, the original deck has $4\\cdot3=\\boxed{12}$ cards.", "Suppose there were $x$ cards in the deck originally. Now, the deck has $x+4$ cards, which must be a multiple of $4.$\nOnly $12+4=16$ is a multiple of $4,$ so the answer is $x=\\boxed{12}.$" ]

[ "There are ${38 \\choose 2} = 703$ ways we can draw two cards from the reduced deck. The two cards will form a pair if both are one of the nine numbers that were not removed, which can happen in $9{4 \\choose 2} = 54$ ways, or if the two cards are the remaining two cards of the number that was removed, which can happen in $1$ way. Thus, the answer is $\\frac{54+1}{703} = \\frac{55}{703}$ , and $m+n = \\boxed{758}$" ]

[ "We can set a proportion:\n\\[\\dfrac{AD}{AB}=\\dfrac{3}{2}\\]\nWe substitute $AB$ with 30 and solve for $AD$\n\\[\\dfrac{AD}{30}=\\dfrac{3}{2}\\]\n\\[AD=45\\]\nWe calculate the combined area of semicircle by putting together semicircle $AB$ and $CD$ to get a circle with radius $15$ . Thus, the area is $225\\pi$ . The area of the rectangle is $30\\cdot 45=1350$ . We calculate the ratio:\n\\[\\dfrac{1350}{225\\pi}=\\dfrac{6}{\\pi}\\Rightarrow\\boxed{6}\\]" ]

[ "Desserts must be chosen for $7$ days: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday.\nThere are $3$ choices for dessert on Saturday: pie, ice cream, or pudding, as there must be cake on Friday and the same dessert may not be served two days in a row. Likewise, there are $3$ choices for dessert on Thursday. Once dessert for Thursday is selected, there are $3$ choices for dessert on Wednesday, once Wednesday's dessert is selected there are $3$ choices for dessert on Tuesday, etc. Thus, there are $3$ choices for dessert for each of $6$ days, so the total number of possible dessert menus is $3^6$ , or $\\boxed{729}$" ]

[ "There are $4 \\cdot 3^6$ ways for the desserts to be chosen. By symmetry, any of the desserts that are chosen on Friday share $\\frac{1}{4}$ of the total arrangements. Therefore our answer is $\\frac{4\\cdot3^6}{4} = 3^6 = \\boxed{729}.$" ]

[ "We can count the number of possible foods for each day and then multiply to enumerate the number of combinations.\nOn Friday, we have one possibility: cake.\nOn Saturday, we have three possibilities: pie, ice cream, or pudding. This is the end of the week.\nOn Thursday, we have three possibilities: pie, ice cream, or pudding. We can't have cake because we have to have cake the following day, which is the Friday with the birthday party.\nOn Wednesday, we have three possibilities: cake, plus the two things that were not eaten on Thursday.\nSimilarly, on Tuesday, we have three possibilities: the three things that were not eaten on Wednesday.\nLikewise on Monday: three possibilities, the three things that were not eaten on Tuesday.\nOn Sunday, it is tempting to think there are four possibilities, but remember that cake must be served on Friday. This serves to limit the number of foods we can eat on Sunday, with the result being that there are three possibilities: The three things that were not eaten on Monday.\nSo the number of menus is $3 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 1 \\cdot 3 = 729.$ The answer is $\\boxed{729}$", "We can perform casework as an understandable means of getting the answer. We can organize our counting based on the food that was served on Wednesday, because whether cake is or is not served on Wednesday seems to significantly affect the number of ways the chef can make said foods for that week.\nCase 1: Cake is served on Wednesday. Here, we have three choices for food on Thursday and Saturday since cake must be served on Friday, and none of these choices are cake, which was served Wednesday. Likewise, we have three choices (pie, ice cream, and pudding) for the food served on Tuesday and thus three choices for those served on Monday and Sunday, with these three choices being whatever was not served on Tuesday and Monday, respectively. Hence, for this case, there are $3 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 3 = 243$ possibilities.\nCase 2: Cake is not served on Wednesday. Obviously, this means that pie, ice cream, and pudding are our only choices for Wednesday's food. Since cake must be served on Friday, only ice cream, pudding, and cake can be served on Thursday. However, since one of those foods was chosen for Wednesday, we only have two possibilities for Thursday's food. Like our first case, we have three possibilities for the food served on Tuesday, Monday, and Sunday: whatever was not served on Wednesday, Tuesday, and Monday, respectively. $3 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 2 \\cdot 3 = 486$ possibilities thus exist for this case.\nAdding the number of possibilities together yields that $243 + 486 = 729$ is the total number of menus, making our answer $\\boxed{729}$", "Note that the choice of a food item on a given day is symmetric, i.e. the number of ways to create the meal plan with a cake on Friday is the same as the number of ways to create the meal plan with pudding on Friday, and the same reasoning holds for the other desserts. Since every meal plan is counted by the summation of the $4$ aforementioned plans (note that Friday's dessert has to be one of the $4$ given desserts) and that these cases are mutually exclusive (i.e. you cannot make both a cake and pudding on Friday), each case results in a quarter of the total meal plans (Since there are $4$ desserts, we multiply by $\\tfrac{1}{4}$ ). The total number of plans with no restrictions can be counted with constructive counting, as follows:\nWe note that there are $4$ ways to choose the first dessert. Then, each dessert thereafter must be distinct from the prior one. Since there are $4$ options, and $1$ of them has been taken by the prior, each following dessert can be chosen in $(4 - 1) = 3$ ways. Thus, since there are $6$ desserts other than the first, the total number (without restrictions) is $4 \\cdot 3^6$\nThus, by our symmetry argument derived prior, we know that the number of desired meal plans is $\\frac{\\text{\\# plans w/o restrictions}}{4} = \\frac{4 \\cdot 3^6}{4} = 3^6 = 729$ , choice $\\boxed{729}$" ]

[ "Do careful casework by each month. In the month and the date, we need a $0$ , a $3$ , and two digits repeated (which has to be $1$ and $2$ after consideration). After the casework, we get $\\boxed{9}$ .\nFor curious readers, the numbers (in chronological order) are:", "There is one $3$ , so we need one more (three more means that either the month or units digit of the day is $3$ ). For the same reason, we need one more $0$\nIf $3$ is the units digit of the month, then the $0$ can be in either of the three remaining slots. For the first case (tens digit of the month), then the last two digits must match ( $11, 22$ ). For the second (tens digit of the day), we must have the other two be $1$ , as a month can't start with $2$ or $0$ . There are $3$ successes this way.\nIf $3$ is the tens digit of the day, then $0$ can be either the tens digit of the month or the units digit of the day. For the first case, $1$ must go in the other slots. For the second, the other two slots must be $1$ as well. There are $2$ successes here.\nIf $3$ is the units digit of the day, then $0$ could go in any of the $3$ remaining slots again. If it's the tens digit of the day, then the other digits must be $1$ . If $0$ is the units digit of the day, then the other two slots must both be $1$ . If $0$ is the tens digit of the month, then the other two slots can be either both $1$ or both $2$ . In total, there are $4$ successes here.\nSumming through all cases, there are $3 + 2 + 4 = \\boxed{9}$ dates.", "We start with $2023----$ we need an extra $0$ and an extra $3$ . So we have at least one of those extras in the days, except we can have the month $03$ . We now have $6$ possible months $01,02,03,10,11,12$ . For month $1$ we have two cases, we now have to add in another 1, and the possible days are $13,31$ . For month $2$ we need an extra $2$ so we can have the day $23$ note that we can't use $32$ because it is to large. Now for month $3$ we can have any number and multiply it by $11$ so we have the solution $11,22$ . For October we need a $1$ and a $3$ so we have $13,31$ as our choices. For November we have two choices which are $03,30$ .Now for December we have $0$ options. Summing $2+1+2+2+2$ we get $\\boxed{9}$ solutions." ]

[ "Do careful casework by each month. In the month and the date, we need a $0$ , a $3$ , and two digits repeated (which has to be $1$ and $2$ after consideration). After the casework, we get $\\boxed{9}$ .\nFor curious readers, the numbers (in chronological order) are:", "There is one $3$ , so we need one more (three more means that either the month or units digit of the day is $3$ ). For the same reason, we need one more $0$\nIf $3$ is the units digit of the month, then the $0$ can be in either of the three remaining slots. For the first case (tens digit of the month), then the last two digits must match ( $11, 22$ ). For the second (tens digit of the day), we must have the other two be $1$ , as a month can't start with $2$ or $0$ . There are $3$ successes this way.\nIf $3$ is the tens digit of the day, then $0$ can be either the tens digit of the month or the units digit of the day. For the first case, $1$ must go in the other slots. For the second, the other two slots must be $1$ as well. There are $2$ successes here.\nIf $3$ is the units digit of the day, then $0$ could go in any of the $3$ remaining slots again. If it's the tens digit of the day, then the other digits must be $1$ . If $0$ is the units digit of the day, then the other two slots must both be $1$ . If $0$ is the tens digit of the month, then the other two slots can be either both $1$ or both $2$ . In total, there are $4$ successes here.\nSumming through all cases, there are $3 + 2 + 4 = \\boxed{9}$ dates.", "We start with $2023----$ we need an extra $0$ and an extra $3$ . So we have at least one of those extras in the days, except we can have the month $03$ . We now have $6$ possible months $01,02,03,10,11,12$ . For month $1$ we have two cases, we now have to add in another 1, and the possible days are $13,31$ . For month $2$ we need an extra $2$ so we can have the day $23$ note that we can't use $32$ because it is to large. Now for month $3$ we can have any number and multiply it by $11$ so we have the solution $11,22$ . For October we need a $1$ and a $3$ so we have $13,31$ as our choices. For November we have two choices which are $03,30$ .Now for December we have $0$ options. Summing $2+1+2+2+2$ we get $\\boxed{9}$ solutions." ]

[ "From the greedy algorithm , we have $9$ in the hours section and $59$ in the minutes section. $9+5+9=\\boxed{23}$", "With a matrix, we can see $\\begin{bmatrix} 1+2&9&6&3\\\\ 1+1&8&5&2\\\\ 1+0&7&4&1 \\end{bmatrix}$ The largest single digit sum we can get is $9$ .\nFor the minutes digits, we can combine the largest $2$ digits, which are $9,5 \\Rightarrow 9+5=14$ , and finally $14+9=\\boxed{23}$", "We first note that since the watch displays time in AM and PM, the value for the hours section varies from $00-12$ . Therefore, the maximum value of the digits for the hours is when the watch displays $09$ , which gives us $0+9=9$\nNext, we look at the value of the minutes section, which varies from $00-59$ . Let this value be a number $ab$ . We quickly find that the maximum value for $a$ and $b$ is respectively $5$ and $9$\nAdding these up, we get $9+5+9=\\boxed{23}$" ]

[ "From the greedy algorithm , we have $9$ in the hours section and $59$ in the minutes section. $9+5+9=\\boxed{23}$", "With a matrix, we can see $\\begin{bmatrix} 1+2&9&6&3\\\\ 1+1&8&5&2\\\\ 1+0&7&4&1 \\end{bmatrix}$ The largest single digit sum we can get is $9$ .\nFor the minutes digits, we can combine the largest $2$ digits, which are $9,5 \\Rightarrow 9+5=14$ , and finally $14+9=\\boxed{23}$", "We first note that since the watch displays time in AM and PM, the value for the hours section varies from $00-12$ . Therefore, the maximum value of the digits for the hours is when the watch displays $09$ , which gives us $0+9=9$\nNext, we look at the value of the minutes section, which varies from $00-59$ . Let this value be a number $ab$ . We quickly find that the maximum value for $a$ and $b$ is respectively $5$ and $9$\nAdding these up, we get $9+5+9=\\boxed{23}$" ]

[ "The center of dilation must lie on the line $A A'$ , which can be expressed as $y = \\dfrac{4x}{3} - \\dfrac{2}{3}$ . Note that the center of dilation must have an $x$ -coordinate less than $2$ ; if the $x$ -coordinate were otherwise, then the circle under the transformation would not have an increased $x$ -coordinate in the coordinate plane. Also, the ratio of dilation must be equal to $\\dfrac{3}{2}$ , which is the ratio of the radii of the circles. Thus, we are looking for a point $(x,y)$ such that $\\dfrac{3}{2} \\left( 2 - x \\right) = 5 - x$ (for the $x$ -coordinates), and $\\dfrac{3}{2} \\left( 2 - y \\right) = 6 - y$ . We do not have to include absolute value symbols because we know that the center of dilation has a lower $x$ -coordinate, and hence a lower $y$ -coordinate, from our reasoning above. Solving the two equations, we get $x = -4$ and $y = - 6$ . This means that any point $(a,b)$ on the plane will dilate to the point $\\left( \\dfrac{3}{2} (a + 4) - 4, \\dfrac{3}{2} (b + 6) - 6 \\right)$ , which means that the point $(0,0)$ dilates to $\\left( 6 - 4, 9 - 6 \\right) = (2,3)$ . Thus, the origin moves $\\sqrt{2^2 + 3^2} = \\boxed{13}$ units.", " Using analytic geometry, we find that the center of dilation is at $(-4,-6)$ and the coefficient/factor is $1.5$ . Then, we see that the origin is $2\\sqrt{13}$ from the center, and will be $1.5 \\times 2\\sqrt{13} = 3\\sqrt{13}$ from it afterwards.\nThus, it will move $3\\sqrt{13} - 2\\sqrt{13} = \\boxed{13}$", "Using the ratios of radii of the circles, $\\frac{3}{2}$ , we find that the scale factor is $1.5$ . If the origin had not moved, this indicates that the center of the\ncircle would be $(3,3)$ , simply because of $(2 \\cdot 1.5, 2 \\cdot 1.5)$ . Since the center has moved from $(3,3)$ to $(5,6)$ , we apply the distance formula and get: $\\sqrt{(6-3)^2 + (5-3)^2} = \\boxed{13}$", "Before dilation, notice that the two axes are tangent to the circle with center $(2,2)$ . Using this, we can draw new axes tangent to the radius 3 circle with center $(5,6)$ , resulting in a \"new origin\" that is 3 units left and 3 units down from the center $(5,6)$ , or $(2,3)$ . Using the distance formula, the distance from $(0,0)$ and $(2,3)$ is $\\boxed{13}$ .\n~Mightyeagle", "Because the dilation changes the circle of radius 2 to a circle of radius 3, the scale factor must be $\\frac{3}{2}$ . The center of the circle with radius 3 is 3 units to the right and 4 units above the center of the circle with radius 2, so the center of dilation must be 6 units to the left and 8 units below the center of the radius 2 circle. So, the center of dilation is the point $(-4, -6)$ . The distance from that point to the origin is $2\\sqrt{13}$ , and halving that to get the distance the origin moves gives $\\boxed{13}$ . ~Meeshbot" ]

[ "The side length of the inner square traced out by the disk with radius $1$ is $s-4.$ However, there is a piece at each corner (bounded by two line segments and one $90^\\circ$ arc) where the disk never sweeps out. The combined area of these four pieces is $(1+1)^2-\\pi\\cdot1^2=4-\\pi.$ As a result, we have \\[A=s^2-(s-4)^2-(4-\\pi)=8s-20+\\pi.\\] Now, we consider the second disk. The part it sweeps is comprised of four quarter circles with radius $2$ and four rectangles with side lengths of $2$ and $s.$ When we add it all together, we have $2A=8s+4\\pi,$ or \\[A=4s+2\\pi.\\] We equate the expressions for $A,$ and then solve for $s:$ \\[8s-20+\\pi=4s+2\\pi.\\] We get $s=5+\\frac{\\pi}{4},$ so the answer is $5+1+4=\\boxed{10}.$" ]

[ "\nLet $F$ be the new tangency point of the two disks. The smaller disk rolled along minor arc $\\overarc{AF}$ on the larger disk.\nLet $\\alpha = \\angle AEF$ , in radians. The smaller disk must then have rolled along an arc of length $5\\alpha$ , since the larger disk has a radius of $5$ . Since all of the points on major arc $\\overarc{BF}$ on the smaller disk have come into contact with the larger disk at some point during the rolling, and none of the other points on the smaller disk did, \\[\\overarc{BF}=\\overarc{AF}=5\\alpha\\]\nSince $\\overline{AC} || \\overline{BD}$ \\[\\angle BDF \\cong \\angle FEA\\] so the angles of minor arc $\\overarc{BF}$ and minor arc $\\overarc{AF}$ are equal, so minor arc $\\overarc{BF}$ has an angle of $\\alpha$ . Since the smaller disk has a radius of $1$ , the length of minor arc $\\overarc{BF}$ is $\\alpha$ . This means that $5\\alpha + \\alpha$ equals the circumference of the smaller disk, so $6\\alpha = 2\\pi$ , or $\\alpha = \\frac{\\pi}{3}$\nNow, to find $\\sin^2{\\angle BEA}$ , we construct $\\triangle BDE$ . Also, drop a perpendicular from $D$ to $\\overline{EA}$ , and call this point $X$ . Since $\\alpha = \\frac{\\pi}{3}$ and $\\angle DXE$ is right, \\[DE = 6\\] \\[EX = 3\\] and \\[DX = 3\\sqrt{3}\\]\nNow drop a perpendicular from $B$ to $\\overline{EA}$ , and call this point $Y$ . Since $\\overline{BD} || \\overline{EA}$ \\[XY = BD = 1\\] and \\[BY = DX = 3\\sqrt{3}\\] Thus, we know that \\[EY = EX - XY = 3 - 1 = 2\\] and by using the Pythagorean Theorem on $\\triangle BEY$ , we get that \\[BE = \\sqrt{31}\\] Thus, \\[\\sin{\\angle BEA} = \\frac{\\sqrt{27}}{\\sqrt{31}}\\] so \\[\\sin^2{\\angle BEA} = \\frac{27}{31}\\] and our answer is $27 + 31 = \\boxed{058}$", "First, we determine how far the small circle goes. For the small circle to rotate completely around the circumference, it must rotate $5$ times (the circumference of the small circle is $2\\pi$ while the larger one has a circumference of $10\\pi$ ) plus the extra rotation the circle gets for rotating around the circle, for a total of $6$ times. Therefore, one rotation will bring point $D$ $60^\\circ$ from $C$\nNow, draw $\\triangle DBE$ , and call \\[\\angle BED = x^{\\circ}\\] We know that $\\overline{ED}$ is 6, and $\\overline{BD}$ is 1. Since $EC || BD$ \\[\\angle BDE = 60^\\circ\\]\nBy the Law of Cosines, \\[\\overline{BE}^2=36+1-2\\times 6\\times 1\\times \\cos{60^\\circ} = 36+1-6=31\\] and since lengths are positive, \\[\\overline{BE}=\\sqrt{31}\\]\nBy the Law of Sines, we know that \\[\\frac{1}{\\sin{x}}=\\frac{\\sqrt{31}}{\\sin{60^\\circ}}\\] so \\[\\sin{x} = \\frac{\\sin{60^\\circ}}{\\sqrt{31}} = \\frac{\\sqrt{93}}{62}\\] As $x$ is clearly between $0$ and $90^\\circ$ $\\cos{x}$ is positive. As $\\cos{x}=\\sqrt{1-\\sin^2{x}}$ \\[\\cos{x} = \\frac{11\\sqrt{31}}{62}\\]\nNow we use the angle sum formula to find the sine of $\\angle BEA$ \\[\\sin 60^\\circ\\cos x + \\cos 60^\\circ\\sin x = \\frac{\\sqrt{3}}{2}\\frac{11\\sqrt{31}}{62}+\\frac{1}{2}\\frac{\\sqrt{93}}{62}\\] \\[= \\frac{11\\sqrt{93}+\\sqrt{93}}{124} = \\frac{12\\sqrt{93}}{124} = \\frac{3\\sqrt{93}}{31} = \\frac{3\\sqrt{31}\\sqrt{3}}{31} = \\frac{3\\sqrt{3}}{\\sqrt{31}}\\]\nFinally, we square this to get \\[\\frac{9\\times 3}{31}=\\frac{27}{31}\\] so our answer is $27+31=\\boxed{058}$" ]

[ "Let $r$ and $b$ denote the number of red and blue socks, respectively. Also, let $t=r+b$ . The probability $P$ that when two socks are drawn randomly, without replacement, both are red or both are blue is given by\n\\[\\frac{r(r-1)}{(r+b)(r+b-1)}+\\frac{b(b-1)}{(r+b)(r+b-1)}=\\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\\frac{1}{2}.\\]\nSolving the resulting quadratic equation $r^{2}-rt+t(t-1)/4=0$ , for $r$ in terms of $t$ , one obtains that\n\\[r=\\frac{t\\pm\\sqrt{t}}{2}\\, .\\]\nNow, since $r$ and $t$ are positive integers, it must be the case that $t=n^{2}$ , with $n\\in\\mathbb{N}$ . Hence, $r=n(n\\pm 1)/2$ would correspond to the general solution. For the present case $t\\leq 1991$ , and so one easily finds that $n=44$ is the largest possible integer satisfying the problem conditions.\nIn summary, the solution is that the maximum number of red socks is $r=\\boxed{990}$", "Let $r$ and $b$ denote the number of red and blue socks such that $r+b\\le1991$ . Then by complementary counting, the number of ways to get a red and a blue sock must be equal to $1-\\frac12=\\frac12=\\frac{2rb}{(r+b)(r+b-1)}\\implies4rb=(r+b)(r+b-1)$ $=(r+b)^2-(r+b)\\implies r^2+2rb+b^2-r-b=4rb\\implies r^2-2rb+b^2$ $=(r-b)^2=r+b$ , so $r+b$ must be a perfect square $k^2$ . Clearly, $r=\\frac{k^2+k}2$ , so the larger $k$ , the larger $r$ $k^2=44^2$ is the largest perfect square below $1991$ , and our answer is $\\frac{44^2+44}2=\\frac12\\cdot44(44+1)=22\\cdot45=11\\cdot90=\\boxed{990}$", "Let $r$ and $b$ denote the number of red and blue socks, respectively. In addition, let $t = r + b$ , the total number of socks in the drawer.\nFrom the problem, it is clear that $\\frac{r(r-1)}{t(t-1)} + \\frac{b(b-1)}{t(t-1)} = \\frac{1}{2}$\nExpanding, we get $\\frac{r^2 + b^2 - r - b}{t^2 - t} = \\frac{1}{2}$\nSubstituting $t$ for $r + b$ and cross multiplying, we get $2r^2 + 2b^2 - 2r - 2b = r^2 + 2br + b^2 - r - b$\nCombining terms, we get $b^2 - 2br + r^2 - b - r = 0$\nTo make this expression factorable, we add $2r$ to both sides, resulting in $(b - r)^2 - 1(b-r) = (b - r - 1)(b - r) = 2r$\nFrom this equation, we can test values for the expression $(b - r - 1)(b - r)$ , which is the multiplication of two consecutive integers, until we find the highest value of $b$ or $r$ such that $b + r \\leq 1991$\nBy testing $(b - r - 1) = 43$ and $(b - r) = 44$ , we get that $r = 43(22) = 946$ and $b = 990$ . Testing values one integer higher, we get that $r = 990$ and $b = 1035$ . Since $990 + 1035 = 2025$ is greater than $1991$ , we conclude that $(946, 990)$ is our answer.\nSince it doesn't matter whether the number of blue or red socks is $990$ , we take the lower value for $r$ , thus the maximum number of red socks is $r=\\boxed{990}$", "As above, let $r$ $b$ , and $t$ denote the number of red socks, the number of blue socks, and the total number of socks, respectively. We see that $\\frac{r(r-1)}{t(t-1)}+\\frac{b(b-1)}{t(t-1)}=\\frac{1}{2}$ , so $r^2+b^2-r-b=\\frac{t(t-1)}{2}=r^2+b^2-t=\\frac{t^2}{2}-\\frac{t}{2}$\nSeeing that we can rewrite $r^2+b^2$ as $(r+b)^2-2rb$ , and remembering that $r+b=t$ , we have $\\frac{t^2}{2}-\\frac{t}{2}=t^2-2rb-t$ , so $2rb=\\frac{t^2}{2}-\\frac{t}{2}$ , which equals $r^2+b^2-t$\nWe now have $r^2+b^2-t=2rb$ , so $r^2-2rb+b^2=t$ and $r-b=\\pm\\sqrt{t}$ . Adding this to $r+b=t$ , we have $2r=t\\pm\\sqrt{t}$ . To maximize $r$ , we must use the positive square root and maximize $t$ . The largest possible value of $t$ is the largest perfect square less than 1991, which is $1936=44^2$ . Therefore, $r=\\frac{t+\\sqrt{t}}{2}=\\frac{1936+44}{2}=\\boxed{990}$", "Let $r$ be the number of socks that are red, and $t$ be the total number of socks. We get:\n$2(r(r-1)+(t-r)(t-r-1))=t(t-1)$ Expanding the left hand side and the right hand side, we get: $4r^2-4rt+2t^2-2t = t^2-t$\nAnd, moving terms, we will get that: $4r^2-4rt+t^2 = t$\nWe notice that the left side is a perfect square. $(2r-t)^2 = t$\nThus $t$ is a perfect square. And, the higher $t$ is, the higher $r$ will be. So, we should set $t = 44^2 = 1936$\nAnd, we see, $2r-1936 = \\pm44$ We will use the positive root, to get that $2r-1936 = 44$ $2r = 1980$ , and $r = \\boxed{990}$", "Let $r$ and $b$ denote the red socks and blue socks, respectively. Thus the equation in question is:\n$\\frac{r(r-1)+b(b-1)}{(r+b)(r+b-1)}=\\frac{1}{2}$\n$\\Rightarrow 2r^2-2r+2b^2-2b=r^2+2rb+b^2-r-b$\n$\\Rightarrow r^2+b^2-r-b-2rb=0$\n$\\Rightarrow (r-b)^2=r+b\\le 1991$\nBecause we wish to maximize $r$ , we have $r\\ge b$ and thus $r-b\\le 44$ as both $r$ and $b$ must be integers and ${44}^2=1936$ is the largest square less than or equal to $1991$ . We now know that the maximum difference between the number of socks is $44$ . Now we return to an earlier equation:\n$r^2+b^2-r-b-2rb=0$\n$\\Rightarrow r^2-(2b+1)r+(b^2-b)=0$\nSolving by the Quadratic formula , we have:\n$r=\\frac{2b+1\\pm\\sqrt{4b^2+4b+1-4b^2+4b}}{2}$\n$\\Rightarrow r=\\frac{2b+1\\pm\\sqrt{8b+1}}{2}$\n$\\Rightarrow r-b=\\frac{1\\pm\\sqrt{8b+1}}{2}$\n$\\Rightarrow 44\\ge r-b=\\frac{1\\pm\\sqrt{8b+1}}{2}$\n$\\Rightarrow 44\\ge\\frac{1\\pm\\sqrt{8b+1}}{2}$\n$\\Rightarrow \\pm\\sqrt{8b+1}\\le 87$\n$\\Rightarrow b\\le 946$ (Here we use the positive sign to maximize $r$ .)\nThus the optimal case would be $b=946$ as $r$ increases only when $b$ increases. In addition, $\\sqrt{8b+1}=87$ as we are using the equality case. We then plug back in:\n$r=\\frac{2b+1\\pm\\sqrt{8b+1}}{2}$\n$\\Rightarrow r=\\frac{1893+87}{2}$ (using the established $\\sqrt{8b+1}=87$\n$\\Rightarrow r=\\frac{1980}{2}$\n$\\Rightarrow r=\\boxed{990}$" ]

[ "After you draw $4$ socks, you can have one of each color, so (according to the pigeonhole principle ), if you pull $\\boxed{5}$ then you will be guaranteed a matching pair." ]

[ "Solution by e_power_pi_times_i\nSuppose that you wish to draw one pair of socks from the drawer. Then you would pick $5$ socks (one of each kind, plus one). Notice that in the worst possible situation, you will continue to draw the same sock, until you get $10$ pairs. This is because drawing the same sock results in a pair every $2$ of that sock, whereas drawing another sock creates another pair. Thus the answer is $5+2\\cdot(10-1) = \\boxed{23}$" ]

[ "After the price reduction, the sale price is $80-.25\\times 80 = 60$ dollars. The tax makes the final price $60+.1\\times 60 = 66$ dollars $\\rightarrow \\boxed{66}$" ]

[ "Since the driver travels $60$ miles per hour and each hour she uses $2$ gallons of gasoline, she spends $$4$ per hour on gas. If she gets $$0.50$ per mile, then she gets $$30$ per hour of driving. Subtracting the gas cost, her net rate of money earned per hour is $\\boxed{26}$ .\n~mathsmiley", "The driver is driving for $2$ hours at $60$ miles per hour, she drives $120$ miles. Therefore, she uses $\\frac{120}{30}=4$ gallons of gasoline. So, total she has $$0.50\\cdot120-$2.00\\cdot4=$60-$8=$52$ . So, her rate is $\\frac{52}{2}=\\boxed{26}$ .\n~sosiaops" ]

[ "Since the driver travels $60$ miles per hour and each hour she uses $2$ gallons of gasoline, she spends $$4$ per hour on gas. If she gets $$0.50$ per mile, then she gets $$30$ per hour of driving. Subtracting the gas cost, her net rate of money earned per hour is $\\boxed{26}$ .\n~mathsmiley", "The driver is driving for $2$ hours at $60$ miles per hour, she drives $120$ miles. Therefore, she uses $\\frac{120}{30}=4$ gallons of gasoline. So, total she has $$0.50\\cdot120-$2.00\\cdot4=$60-$8=$52$ . So, her rate is $\\frac{52}{2}=\\boxed{26}$ .\n~sosiaops" ]

[ "Recast the problem entirely as a block-walking problem. Call the respective dice $a, b, c, d$ . In the diagram below, the lowest $y$ -coordinate at each of $a$ $b$ $c$ , and $d$ corresponds to the value of the roll.\nAIME01IN6.png\nThe red path corresponds to the sequence of rolls $2, 3, 5, 5$ . This establishes a bijection between valid dice roll sequences and block walking paths.\nThe solution to this problem is therefore $\\dfrac{\\binom{9}{4}}{6^4} = {\\dfrac{7}{72}}$ . So the answer is $\\boxed{079}$", "Let $a, b, c,$ and $d$ be the results of rolling the four dice respectively. We have the range $1\\leq a\\leq b\\leq c\\leq d\\leq 6$ , and there are $6^4=1296$ total outcomes from rolling the dice. To transfer the inequality into a strictly increasing inequality, we can transform it into $1\\leq a<b+1<c+2<d+3\\leq 9$ . Now, lets suppose $a'=a, b'=b+1, c'=c+2,$ and $d'=d+3$ . Then, $1\\leq a'<b'<c'<d' \\leq 9$ . Clearly, there are $\\binom{9}{4}=126$ values that satisfy this equation, so our answer is $\\dfrac{\\binom{9}{4}}{6^4} = {\\dfrac{7}{72}}\\implies \\boxed{079}$", "If we take any combination of four numbers, there is only one way to order them in a non-decreasing order. It suffices now to find the number of combinations for four numbers from $\\{1,2,3,4,5,6\\}$ . We can visualize this as taking the four dice and splitting them into 6 slots (each slot representing one of {1,2,3,4,5,6}), or dividing them amongst 5 separators. Thus, there are ${9\\choose4} = 126$ outcomes of four dice. The solution is therefore $\\frac{126}{6^4} = \\frac{7}{72}$ , and $7 + 72 = \\boxed{079}$", "Call the dice rolls $a, b, c, d$ . The difference between the $a$ and $d$ distinguishes the number of possible rolls there are.\nContinuing, we see that the sum is equal to $\\sum_{i = 0}^{5}{i+2\\choose2}(6 - i) = 6 + 15 + 24 + 30 + 30 + 21 = 126$ . The requested probability is $\\frac{126}{6^4} = \\frac{7}{72}$ and our answer is $\\boxed{79}$", "The dice rolls can be in the form\nwhere A, B, C, D are some possible value of the dice rolls. (These forms are not keeping track of whether or not the dice are in ascending order, just the possible outcomes.)\nThere are a total of $6^4$ possible dice rolls.\nThus,\nTherefore, our answer is $\\boxed{079}$", "In a manner similar to the above solution, instead consider breaking it down into two sets of two dice rolls. The first subset must have a maximum value which is $\\le$ the minimum value of the second subset.\nThus, the number of combinations is $\\sum_{i=1}^6 i \\cdot \\left(\\frac{(7 - i)(8 - i)}{2}\\right) = 21 + 30 + 30 + 24 + 15 + 6 = 126$ , and the probability again is $\\frac7{72}$ , giving $m+n=\\boxed{079}$", "Lets try casework and observe the cases. Notice that if the last roll is a $1$ , then the only dice rolls may be $1-1-1-1$ , which is only $1$ possibility. Observe that if the last roll is $2$ , then there are $4 = 1 + 3$ possibilities. When the last roll is a $3$ , there are $10 = 1 + 3 + 6$ possibilities. Notice when the last roll is $n$ , the number of cases is the sum of the first $n$ positive triangular numbers. This is easily provable when observing the numbers of possibilities after assigning a value to the last and second to last rolls.\nSo there are a total of $1 + 1 + 3 + 1 + 3 + 6 + 1 + 3 + 6 + 10 + 1 + 3 + 6 + 10 + 15 + 1 + 3 + 6 + 10 + 15 + 21 = 126$ possibilities. So the probability is $\\frac{126}{6^4} = \\frac{7}{72}$ and $7 + 72 = \\boxed{079}$ . ~skyscraper", "This is equivalent to picking a four-element sequence of $\\{1, 2, 3, 4, 5, 6\\}$ with repetition. Notice that once the sequence is picked, there is one and only one way to order these so that they form a sequence of rolls satisfying our conditions.\nNow count the number of such four-element sequences, let $a$ be the number of $1$ s in the sequence, $b$ be the number of $2$ s, $c$ $3$ s, $d$ $4$ s, $e$ $5$ s, and $f$ $6$ s. Now we see that we must have \\[a + b + c + d + e + f = 4\\] with $a, b, c, d, e, f$ being nonnegative integers since there are a total of $4$ numbers picked. The number of solutions to this is $\\dbinom{9}{4},$ so our total number is equal to $\\dfrac{\\binom{9}{4}}{6^4} = \\dfrac{7}{72},$ making our answer $\\boxed{079}.$", "Let the rolls be $a,b,c$ and $d\\newline$ let $z=a-1, e=b-a, f=c-b, g=d-c, h=6-d\\newline$ $z+e+f+g+h=5\\newline$ This equation has $C(5+5-1, 5-1)=126$ integer solutions $\\newline$ $126/1296=7/72\\newline$ $7+72=\\boxed{79}$ ~ryanbear", "Let's say the four dice values are all different. These can only be arranged in one way to satisfy our conditions, so there are $\\binom{6}{4}=15$ ways. If there are three different values, there are $\\binom{6}{3}$ to choose the numbers and $\\binom{3}{1}$ to choose which number will have the repeat, so $20\\cdot3=60$ ways. If there are two different values, there are $\\binom{6}{2}$ ways to choose the numbers. If there are two of each, then there is one way. If there are three of one and one of another, then there are $\\binom{2}{1}$ ways. Therefore $15\\cdot(1+2)=45$ . Now if all the numbers are the same, there are $\\binom{6}{1}=6$ ways. Altogether we have $15+60+45+6=126$ ways. $\\frac{126}{6^4}=\\frac{21}{216}=\\frac{7}{72}$ $7+72=\\boxed{79}$ . \n~MC413551" ]

[ "We can work in reverse by first determining the number of combinations in which there are more than $2$ friends in at least one room. There are three cases:\nCase 1: Three friends are in one room. Since there are $5$ possible rooms in which this can occur, we are choosing three friends from the five, and the other two friends can each be in any of the four remaining rooms,\nthere are $5\\cdot\\binom{5}{3}\\cdot4\\cdot4 = 800$ possibilities.\nCase 2: Four friends are in one room. Again, there are $5$ possible rooms, we are choosing four of the five friends, and the other one can be in any of the other four rooms, so there are $5\\cdot\\binom{5}{4}\\cdot4= 100$ possibilities.\nCase 3: Five friends are in one room. There are $5$ possible rooms in which this can occur, so there are $5$ possibilities.\nSince there are $5^5 = 3125$ possible combinations of the friends, the number fitting the given criteria is $3125 - (800+100+5) = \\boxed{2220}$" ]

[ "Let us say each sheep cost $x$ dollars. The farmer paid $749x$ for the sheep. He sold $700$ of them for $749x$ , so each sheep sold for $\\frac{749}{700} = 1.07x$\nSince every sheep sold for the same price per head, and since every sheep cost $x$ and sold for $1.07x$ , there is an increase of $\\frac{1.07x - 1x}{1x} = 0.07$ , or $7\\%$ , which is option $\\boxed{7}$" ]

[ "The first three sons get $\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{5}=\\frac{19}{20}$ of the herd, so that the fourth son should get $\\frac{1}{20}$ of it. But the fourth son gets $7$ cows, so the size of the herd is $n=\\frac{7}{\\frac{1}{20}} = 140$ . Then our answer is $\\boxed{140}$ , and we are done." ]

[ "There are $4$ possibilities for the top-left section. It follows that the top-right and bottom-left sections each have $3$ possibilities, so they have $3^2=9$ combinations. We have two cases:\nTogether, the answer is $36+48=\\boxed{84}.$", "We will do casework on the type of crops in the field.\nCase 1: all of a kind.\nIf all four sections have the same type of crop, there are simply $\\underline{4}$ ways to choose crops for the sections.\nCase 2: $\\boldsymbol{3}$ of a kind, $\\boldsymbol{1}$ of another kind.\nSince the one of another kind must be adjacent to two of the other crops, when choosing the type of crops in this case, we cannot choose soybeans and potatoes, or corn and wheat. Therefore, there are $4 \\cdot 3 - 2 \\cdot 2 = 8$ choices for the two crops we choose for the section (notice we did not choose by $2$ , since the crop we pick first will be the unique one), and $4$ ways to choose which section the unique crop is planted on. This gives us a total of $4 \\cdot 8 = \\underline{32}$ ways to choose crops for the sections.\nCase 3: $\\boldsymbol{2}$ of a kind, $\\boldsymbol{2}$ of another kind.\nWe cannot choose corn and wheat, or soybeans and potatoes, once again, because if we do, the two would have to be adjacent in some way, which the problem disallows. So, there are ${4 \\choose 2} - 2 = 4$ ways to choose our two crops (notice that we did choose by $2$ , since there are two of both crops). There are ${4 \\choose 2} = 6$ ways to choose where one of the crops go, so there are $4 \\cdot 6 = \\underline{24}$ ways to choose crops for the sections.\nCase 4: $\\boldsymbol{2}$ of a kind, $\\boldsymbol{1}$ of another kind, $\\boldsymbol{1}$ of another kind.\nIn cases 2 and 3, we excluded the possibility of choosing bad pairs for our crops (i.e. soybeans and potatoes, or corn and wheat). In this case, it is inevitable that we choose a bad pair, because we are choosing $3$ crops this time. The two sections of the same kind must contain the crop that is not part of the bad pair in the trio: for example, if we choose corn, soybeans and potatoes as our three crop types, nor soybeans and potatoes can be the type which occupies two sections in this case; corn must be the one to do so. There are $4$ ways to choose the crop that is not part of the bad pair, and then $1$ way to choose the bad pair, giving us $4 \\cdot 1 = 4$ ways to choose the crops. To separate the bad pair of crops, the two of a kind must be diagonally placed. There are $2$ ways to choose where the two of a kind go, and $2$ ways to choose which of the bad pair goes where, giving us $2 \\cdot 2 = 4$ ways to choose the positions for the crops. In total, there are $4 \\cdot 4 = \\underline{16}$ ways to choose crops for the sections.\nCase 5: every single crop.\nBad pairs must be on the same diagonal, so there are $2$ ways to choose which pair gets which diagonal, and $2 \\cdot 2 = 4$ ways to choose which of each pair goes where on the diagonal, giving us $2 \\cdot 4 = \\underline{8}$ ways to choose crops for the sections.\nAdding up all our values, we get our final answer of $4+32+24+16+8 = \\boxed{84}$", "To lighten notation, we use C, W, S, P to denote corn, wheat, soybeans, and potatoes, respectively.\nWe use I, II, III, IV to denote four quadrants, respectively.\nWe determine an arrangement in the following steps.\nStep 1: Determine the crop planted in I.\nThe number of ways is $4$\nStep 2: Determine the crops planed in II, III, IV.\nTo find the number of arrangements in this step, without loss of generality, we assume that we plant C in I.\nWe do the following casework analysis.\nCase 1: Both II and IV are planted with C.\nIn this case, the number of ways to plant in III is $3$\nCase 2: In II and IV, only one quadrant is planted with C, and another quadrant is planted with either S or P.\nIn this case, we determine an arrangement in the following steps.\nStep 2.1: Determine whether C is planted in II or IV.\nThe number of ways is $2$\nStep 2.2: In II or IV not planted with C, determine whether it is planted with S or P.\nThe number of ways is $2$\nStep 2.3: Determine the crop planted in III.\nThe number of ways is $2$\nFollowing from the rule of product, the number of ways in this case is $2 \\cdot 2 \\cdot 2 = 8$\nCase 3: II and IV are both planted with S or W.\nIn this case, we determine an arrangement in the following steps.\nStep 2.1: Determine whether S or W is planted in II and IV.\nThe number of ways is $2$\nStep 2.2: Determine the crop planted in III.\nThe number of ways is $3$\nFollowing from the rule of product, the number of ways in this case is $2 \\cdot 3 = 6$\nCase 4: In II and IV, exactly one quadrant is planted with S and another one is planted with W.\nStep 2.1: Determine which quadrant in II and IV is planted with S.\nThe number of ways is $2$\nStep 2.2: Determine the crop planted in III.\nThe number of ways is $2$\nFollowing from the rule of product, the number of ways in this case is $2 \\cdot 2 = 4$\nPutting all cases together, the total number of arrangements in Step 2 is $3 + 8 + 6 + 4 = 21$\nFollowing from the rule of product, the total number of arrangements is $4 \\cdot 21 = 84$\nTherefore, the answer is $\\boxed{84}$", "The number of cases with at least one pair of corn and wheat adjacent is $4 \\cdot 2 \\cdot 4^2 - 4 \\cdot 2 \\cdot 4 + 0 - 2 = 94$ possible fields (You can easily see this for yourself using PIE.), and WLOG, the same goes for soybean and potatoes. Now, applying PIE on both sets (number of cases with at least one pair of corn and wheat and the number of cases with at least one pair of soybeans and potatoes) yields $2\\cdot94 - 16 = 172$ . We want the number of cases without adjacent pairs of soybeans and potatoes or wheat and corn, so we subtract $256 - 172$ , yielding an answer of $\\boxed{84}.$", "We can create a tree representing an arbitrary box we start with and the possibilities for other boxes around the grid. We can just designate the crop we start with as $1$ , and the other crops as $2$ $3$ , and $4$ , where $1$ cannot be next to $2$ and $3$ cannot be next to $4$ in the grid. As we can see, there are $21$ possibilities. We multiply this by the $4$ (the number of possibilities there are for the crop we started with), so our answer is $\\boxed{84}$", "The top right box has $4$ choices and the top left box has $3$ choices. Thus, it is reasonable to assume that the answer is a multiple of $12$ . We know that the answer will not be too small or too large, so the answer is $\\boxed{84}$" ]

[ "The bill for the three children is $9.45 - 4.95 = 4.50$ . Since the charge is $0.45$ per year for the children, they must have $\\frac{4.50}{0.45} = 10$ years among the three of them.\nThe twins must have an even number of years in total (presuming that they did not dine in the 17 minutes between the time when the first twin was born and the second twin was born). If we let the twins be $5$ years old, that leaves $10 - 2\\cdot 5 = 0$ years leftover for the youngest child. But if the twins are each $4$ years old, then the youngest child could be $10 - 2\\cdot 4 = 2$ years old, which is choice $\\boxed{2}$" ]

[ "Alternatively, consider that counting without the number $4$ is almost equivalent to counting in base $9$ ; only, in base $9$ , the number $9$ is not counted. Since $4$ is skipped, the symbol $5$ represents $4$ miles of travel, and we have traveled $2004_9$ miles. By basic conversion, $2004_9=9^3(2)+9^0(4)=729(2)+1(4)=1458+4=\\boxed{1462}$", "This is very analogous to base $9$ . But, in base $9$ , we don't have a $9$ . So, this means that these are equal except for that base 9 will be one more than the operation here. $2005_9 = 5+0+0+1458 = 1463$ $1463 - 1 = 1462$\nTherefore, our answer is $\\boxed{1462}$" ]

[ "Suppose there are $n$ squares in every column of the grid, so there are $\\frac{52}{24}n = \\frac {13}6n$ squares in every row. Then $6|n$ , and our goal is to maximize the value of $n$\nEach vertical fence has length $24$ , and there are $\\frac{13}{6}n - 1$ vertical fences; each horizontal fence has length $52$ , and there are $n-1$ such fences. Then the total length of the internal fencing is $24\\left(\\frac{13n}{6}-1\\right) + 52(n-1) = 104n - 76 \\le 1994 \\Longrightarrow n \\le \\frac{1035}{52} \\approx 19.9$ , so $n \\le 19$ . The largest multiple of $6$ that is $\\le 19$ is $n = 18$ , which we can easily verify works, and the answer is $\\frac{13}{6}n^2 = \\boxed{702}$", "Assume each partitioned square has a side length of $1$ (just so we can get a clear image of what the formula will look like). The amount of internal fencing that is required to partition the field is clearly $52*(24+1) + 24(52+1)$ . (If you are confused, just draw the square out). This is clearly greater than $1994$ , so the actual side length that we are looking for is greater than $1$\nNow we can convert this into an equation. The equation is simply $(\\frac{52}{x})(\\frac{24}{x}+1) + (\\frac{24}{x})(\\frac{52}{x}+1)$ (The intutition comes from considering partioning the field into side lengths of $1$ and then partitioning those squares). This is equivalent to $\\frac{2496}{x^2} + \\frac{76}{x}$ , which should be less than or equal to $1994$\nNow we can just find possible lengths of the square that are greater than $1$ and test them out. A viable side length would mean that $\\frac{24}{x}, \\frac{52}{x} \\in$ N. Since $\\gcd(24,52) = 4$ , then the smallest value greater than $1$ that we satisfies the conditions has $4$ in the numerator, and hence $3$ in the denominator. Test out $x=\\frac{4}{3}$ . This will equate to something less than $1994$ , and hence the smallest square length that is plausible is $\\frac{4}{3}$\nNow the rest is elementary, we do $\\frac{52}{\\frac{4}{3}} * \\frac{24}{\\frac{4}{3}} \\Rightarrow 39*18 = \\boxed{702}$" ]

[ "It is easy to see that the ferry boat takes $6$ trips total. The total number of people taken to the island is\n\\begin{align*}&100+99+98+97+96+95\\\\ &=6(100)-(1+2+3+4+5)\\\\ &=600 - 15\\\\ &=\\boxed{585}" ]

[ "The possible medians after $n$ is added are $6$ $n$ , or $9$ . Now we use casework\nCase 1: The median is $6$\nIn this case, $n<6$ and \\[\\frac{3+n+6+9+10}{5}=6 \\Rightarrow n=2\\] so this case contributes $1$\nCase 2: The median is $n$\nWe have $6<n<9$ and \\[\\frac{3+6+n+9+10}{5}=n \\Rightarrow n=7\\] so this case also contributes $1$\nCase 3: The median is $9$\nWe have $9<n$ and \\[\\frac{3+6+9+n+10}{5}=9 \\Rightarrow 17\\] so this case adds $1$\nIn all there are $3\\rightarrow \\boxed{3}$ possible values of $n$" ]

[ "In order to minimize the amount of cubes needed, we must match up as many squares of our given figures with each other to make different sides of the same cube. One example of the solution with $\\boxed{4}$ cubes. Notice the corner cube cannot be removed for a figure of 3 cubes because each face of a cube must be touching another face." ]

[ "A given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let $k$ be the sum of the units digits in all the terms. Then $S=111k=3 \\cdot 37k$ , so $S$ must be divisible by $37\\ \\mathrm{(D)}$ . To see that it need not be divisible by any larger prime, the sequence $123, 231, 312$ gives $S=666=2 \\cdot 3^2 \\cdot 37\\Rightarrow \\mathrm{\\boxed{37}$" ]

[ "A given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let $k$ be the sum of the units digits in all the terms. Then $S=111k=3 \\cdot 37k$ , so $S$ must be divisible by $37\\ \\mathrm{(D)}$ . To see that it need not be divisible by any larger prime, the sequence $123, 231, 312$ gives $S=666=2 \\cdot 3^2 \\cdot 37\\Rightarrow \\mathrm{\\boxed{37}$" ]

[ "Let $x$ be the mean of $\\mathcal{S}$ . Let $a$ be the number of elements in $\\mathcal{S}$ .\nThen, the given tells us that $\\frac{ax+1}{a+1}=x-13$ and $\\frac{ax+2001}{a+1}=x+27$ . Subtracting, we have \\begin{align*}\\frac{ax+2001}{a+1}-40=\\frac{ax+1}{a+1} \\Longrightarrow \\frac{2000}{a+1}=40 \\Longrightarrow a=49\\end{align*} We plug that into our very first formula, and get: \\begin{align*}\\frac{49x+1}{50}&=x-13 \\\\ 49x+1&=50x-650 \\\\ x&=\\boxed{651}", "Since this is a weighted average problem, the mean of $S$ is $\\frac{13}{27}$ as far from $1$ as it is from $2001$ . Thus, the mean of $S$ is $1 + \\frac{13}{13 + 27}(2001 - 1) = \\boxed{651}$" ]